GATE EC · EE · IN

Laplace transform, poles, zeros and system functions

The Laplace transform turns the differential equation of a continuous-time LTI system into algebra in the complex variable $s = σ + j ω$ , and its region of convergence (ROC) decides which time signal an algebraic $X (s)$ actually stands for. The poles, zeros and ROC of the system function $H (s)$ answer what GATE asks of such a system: its impulse and step responses, whether it is causal and whether it is stable, what its frequency response looks like, and how it responds to initial conditions.

8 previous-year questions · 8 solved · 7 practice questions

Definition, conventions and the region of convergence

For a signal $x (t)$ and a complex variable $s = σ + j ω$ , the bilateral Laplace transform integrates over all time. Writing $e^{- s t} = e^{- σ t} e^{- j ω t}$ shows what it is: the Fourier transform of $x (t) e^{- σ t}$ . The factor $e^{- σ t}$ can tame a growing signal, which is why $e^{2 t} u (t)$ has no Fourier transform but does have a Laplace transform.

The set of $s$ for which the integral converges absolutely is the region of convergence (ROC). An algebraic expression $X (s)$ without its ROC does not define a signal: the same $X (s)$ with different ROCs belongs to different signals.

Conventions in these notes. $j = - 1$ . The Fourier transform is written in $ω$ (rad/s): $X (j ω) = \int_{- \infty}^{\infty} x (t) e^{- j ω t} d t$ , with $\frac{1}{2 π}$ in the inverse. LHP and RHP mean the open left and right halves of the $s$ -plane; the $j ω$ axis belongs to neither. Each result below says whether it is for the bilateral or the unilateral transform, and the ROC it holds in.

BILATERAL LAPLACE TRANSFORM

X (s) = L {x (t)} = \int_{- \infty}^{\infty} x (t) e^{- s t} d t, s = σ + j ω

Converges only for $s$ in the ROC, which must be quoted with $X (s)$ .

UNILATERAL LAPLACE TRANSFORM

X (s) = \int_{0^{-}}^{\infty} x (t) e^{- s t} d t

Lower limit $0^{-}$ , so an impulse at $t = 0$ is included. It equals the bilateral transform of the part of $x (t)$ from $t = 0^{-}$ onwards, so, when it exists, its ROC is a right half-plane $Re (s) > σ_{0}$ (the whole plane if that part has finite duration) and is usually not written. For a causal signal the two transforms are identical. This is the version used for differential equations with initial conditions.

LINK TO THE FOURIER TRANSFORM

X (j ω) = X (s)_{s = j ω} (only if the ROC contains the j ω axis)

In the Hz form, $X (f) = X (s)_{s = j 2 π f}$ . If a pole lies on the $j ω$ axis (as for $u (t)$ ) the substitution is not valid.

INVERSE LAPLACE TRANSFORM

x (t) = \frac{1}{2 π j} \int_{σ - j \infty}^{σ + j \infty} X (s) e^{s t} d s, σ inside the ROC

Not evaluated directly in practice: partial fractions plus the table of pairs, read with the ROC, is the method.

Properties of the ROC

The ROC depends only on $σ = Re (s)$ , so it is made of vertical strips parallel to the $j ω$ axis.
The ROC contains no poles. For a rational $X (s)$ it is bounded by poles or extends to infinity.
The ROC is one connected region: a right half-plane, a left half-plane, a vertical strip, the whole plane, or empty (then there is no Laplace transform).
Finite-duration, absolutely integrable $x (t)$ : the ROC is the entire $s$ -plane.
Right-sided $x (t)$ (zero for $t < T_{1}$ ): if the line $Re (s) = σ_{0}$ is in the ROC, so is every $Re (s) > σ_{0}$ . For rational $X (s)$ the ROC is the half-plane to the right of the rightmost pole.
Left-sided $x (t)$ (zero for $t > T_{2}$ ): the ROC is a left half-plane; for rational $X (s)$ , to the left of the leftmost pole.
Two-sided $x (t)$ : the ROC is a vertical strip (between two adjacent poles, for rational $X (s)$ ) or empty.
A rational $X (s)$ whose poles have $k$ distinct real parts admits $k + 1$ possible ROCs: the half-plane to the left of every pole, the half-plane to the right of every pole, and the $k - 1$ strips between adjacent real parts.
The ROC of a sum contains the intersection of the individual ROCs; it can be larger if a pole cancels.

Signal	$X (s)$	ROC
$e^{- a t} u (t)$ , right-sided ( $a$ real)	$\frac{1}{s + a}$	$Re (s) > - a$
$- e^{- a t} u (- t)$ , left-sided ( $a$ real)	$\frac{1}{s + a}$	$Re (s) < - a$
$e^{- a ∣ t ∣}$ with $a > 0$ , two-sided	$\frac{- 2 a}{s ^{2} - a ^{2}}$	$- a < Re (s) < a$
$u (t) - u (t - T)$ with $T > 0$ , finite duration	$\frac{1 - e ^{- s T}}{s}$	entire $s$ -plane (the pole at $0$ cancels)

WORKED EXAMPLE · SAME X(S), THREE SIGNALS

$X (s) = \frac{3}{( s + 1 ) ( s - 2 )}$ . Find $x (t)$ for every possible ROC and state which of these signals has a Fourier transform.

Solution

Partial fractions: $X (s) = \frac{A}{s + 1} + \frac{B}{s - 2}$ with $A = \frac{3}{s - 2}_{s = - 1} = - 1$ and $B = \frac{3}{s + 1}_{s = 2} = 1$ .
The poles $- 1$ and $2$ allow three ROCs: $Re (s) > 2$ , $- 1 < Re (s) < 2$ and $Re (s) < - 1$ . Rule for each term: a pole to the left of the ROC gives a right-sided term with $u (t)$ ; a pole to the right of the ROC gives a left-sided term with $- u (- t)$ .
$Re (s) > 2$ : both poles are left of the ROC, so $x (t) = (e^{2 t} - e^{- t}) u (t)$ , causal and growing without bound.
$- 1 < Re (s) < 2$ : the pole at $- 1$ is left of the ROC and the pole at $2$ is right of it, so $x (t) = - e^{- t} u (t) - e^{2 t} u (- t)$ , two-sided and decaying in both directions.
$Re (s) < - 1$ : both poles are right of the ROC, so $x (t) = (e^{- t} - e^{2 t}) u (- t)$ , anti-causal.
Only the strip $- 1 < Re (s) < 2$ contains the $j ω$ axis, so only the two-sided signal is absolutely integrable, with $X (j ω) = \frac{3}{( j ω + 1 ) ( j ω - 2 )}$ .

Answer: $Re (s) > 2$ : $(e^{2 t} - e^{- t}) u (t)$ . $- 1 < Re (s) < 2$ : $- e^{- t} u (t) - e^{2 t} u (- t)$ , the only one with a Fourier transform. $Re (s) < - 1$ : $(e^{- t} - e^{2 t}) u (- t)$ .

EXAM TIP

Intersect the ROCs of the pieces before writing $X (s)$ . $x (t) = e^{- 2 t} u (t) + e^{- 3 t} u (- t)$ has piece ROCs $Re (s) > - 2$ and $Re (s) < - 3$ , which do not overlap, so it has no Laplace transform at all. Likewise the everlasting $cos (ω_{0} t)$ (for all $t$ ) has a Fourier transform made of impulses but no bilateral Laplace transform: its right-sided half needs $Re (s) > 0$ and its left-sided half needs $Re (s) < 0$ .

EXAM TIP

Watch the minus sign on left-sided pairs. $e^{- a t} u (t)$ and $- e^{- a t} u (- t)$ have the same algebraic transform $\frac{1}{s + a}$ ; only the ROC and that minus sign tell them apart.

Standard pairs and properties

Almost every transform and inverse transform is built from a handful of right-sided pairs and the properties below. The pairs are written for causal signals, so they are also unilateral transforms. For every rational pair below other than $δ (t)$ , the left-sided signal $- x (t) u (- t)$ has the same algebraic $X (s)$ as the causal $x (t) u (t)$ , with the ROC to the left of the leftmost pole instead of to the right of the rightmost. In the power, exponential and sinusoid pairs all the poles lie on one vertical line $Re (s) = σ_{0}$ , so the two ROCs are simply $Re (s) > σ_{0}$ and $Re (s) < σ_{0}$ .

IMPULSE, STEP, RAMP AND POWERS OF T

δ (t) \leftrightarrow 1 (all s); u (t) \leftrightarrow \frac{1}{s}, t u (t) \leftrightarrow \frac{1}{s ^{2}}, \frac{t ^{n}}{n !} u (t) \leftrightarrow \frac{1}{s ^{n + 1}} (each Re (s) > 0)

$n = 0, 1, 2, \dots$ Also, for all $s$ : $δ (t - t_{0}) \leftrightarrow e^{- s t_{0}}$ , and the derivatives of the impulse $\frac{d ^{n} δ ( t )}{d t ^{n}} \leftrightarrow s^{n}$ (so $δ^{'} (t) \leftrightarrow s$ ).

EXPONENTIALS

e^{- a t} u (t) \leftrightarrow \frac{1}{s + a}, \frac{t ^{n}}{n !} e^{- a t} u (t) \leftrightarrow \frac{1}{( s + a ) ^{n + 1}} (Re (s) > - Re (a))

Valid for real or complex $a$ . The left-sided versions are $- e^{- a t} u (- t)$ and $- \frac{t ^{n}}{n !} e^{- a t} u (- t)$ , with ROC $Re (s) < - Re (a)$ . For real $a$ the hyperbolic pairs follow: $cosh (a t) u (t) \leftrightarrow \frac{s}{s ^{2} - a ^{2}}$ and $sinh (a t) u (t) \leftrightarrow \frac{a}{s ^{2} - a ^{2}}$ , both with ROC $Re (s) > ∣ a ∣$ .

SINUSOIDS

cos (ω_{0} t) u (t) \leftrightarrow \frac{s}{s ^{2} + ω _{0}^{2}}, sin (ω_{0} t) u (t) \leftrightarrow \frac{ω _{0}}{s ^{2} + ω _{0}^{2}} (Re (s) > 0)

$ω_{0}$ real. The poles at $\pm j ω_{0}$ lie on the $j ω$ axis, which the ROC does not contain, so putting $s = j ω$ does not give the Fourier transform of these one-sided sinusoids (theirs contains impulses at $ω = \pm ω_{0}$ ).

SINUSOIDS MULTIPLIED BY T (DOUBLE POLES ON THE JΩ AXIS)

t cos (ω_{0} t) u (t) \leftrightarrow \frac{s ^{2} - ω _{0}^{2}}{( s ^{2} + ω _{0}^{2} ) ^{2}}, t sin (ω_{0} t) u (t) \leftrightarrow \frac{2 ω _{0} s}{( s ^{2} + ω _{0}^{2} ) ^{2}} (Re (s) > 0)

From the sinusoid pairs by multiplication by $t$ . Combining them gives $\frac{1}{( s ^{2} + ω _{0}^{2} ) ^{2}} \leftrightarrow \frac{sin ( ω _{0} t ) - ω _{0} t cos ( ω _{0} t )}{2 ω _{0}^{3}} u (t)$ : a double pole on the $j ω$ axis gives a term that grows like $t$ .

DAMPED SINUSOIDS

e^{- a t} cos (ω_{0} t) u (t) \leftrightarrow \frac{s + a}{( s + a ) ^{2} + ω _{0}^{2}}, e^{- a t} sin (ω_{0} t) u (t) \leftrightarrow \frac{ω _{0}}{( s + a ) ^{2} + ω _{0}^{2}} (Re (s) > - a)

$a$ and $ω_{0}$ real. Poles at $- a \pm j ω_{0}$ . These are the undamped pairs with $s$ replaced by $s + a$ (shift in $s$ ).

CAUSAL PERIODIC SIGNAL

x (t) = n = 0 \sum \infty x_{1} (t - n T) \leftrightarrow X (s) = \frac{X _{1} ( s )}{1 - e ^{- s T}}, Re (s) > 0

$x_{1} (t)$ is one period: absolutely integrable and zero outside $0 \leq t < T$ . Example: the wave that is $1$ for $0 \leq t < T /2$ and $0$ for $T /2 \leq t < T$ , repeated for $t \geq 0$ , has $X (s) = \frac{1 - e ^{- s T /2}}{s ( 1 - e ^{- s T} )} = \frac{1}{s ( 1 + e ^{- s T /2} )}$ .

Property	Time domain	$s$ -domain (bilateral)	ROC
Linearity	$a x_{1} (t) + b x_{2} (t)$	$a X_{1} (s) + b X_{2} (s)$	contains $R_{1} \cap R_{2}$
Time shift	$x (t - t_{0})$	$e^{- s t_{0}} X (s)$	$R$
Shift in $s$	$e^{s_{0} t} x (t)$	$X (s - s_{0})$	$R$ shifted by $Re (s_{0})$
Modulation, $ω_{0}$ real	$x (t) cos (ω_{0} t)$	$\frac{1}{2} [X (s - j ω_{0}) + X (s + j ω_{0})]$	contains $R$
Time scaling, $a \neq = 0$	$x (a t)$	$\frac{1}{∣ a ∣} X (\frac{s}{a})$	$a R$ (flipped if $a < 0$ )
Time reversal	$x (- t)$	$X (- s)$	$- R$
Differentiation in $t$	$\frac{d x ( t )}{d t}$	$s X (s)$	contains $R$
Multiplication by $t$	$t x (t)$	$- \frac{d X ( s )}{d s}$	$R$
Integration	$\int_{- \infty}^{t} x (τ) d τ$	$\frac{X ( s )}{s}$	contains $R \cap {Re (s) > 0}$
Convolution	$x_{1} (t) * x_{2} (t)$	$X_{1} (s) X_{2} (s)$	contains $R_{1} \cap R_{2}$
Conjugation	$x^{*} (t)$	$X^{} (s^{})$	$R$

MULTIPLICATION BY POWERS OF T

t^{n} x (t) \leftrightarrow (- 1)^{n} \frac{d ^{n} X ( s )}{d s ^{n}}

Same ROC as $X (s)$ . With $n = 1$ and $x (t) = e^{- a t} u (t)$ it gives $t e^{- a t} u (t) \leftrightarrow \frac{1}{( s + a ) ^{2}}$ .

UNILATERAL DIFFERENTIATION (BRINGS IN INITIAL CONDITIONS)

\frac{d x}{d t} \leftrightarrow s X (s) - x (0^{-}), \frac{d ^{2} x}{d t ^{2}} \leftrightarrow s^{2} X (s) - s x (0^{-}) - x^{'} (0^{-})

Unilateral transform only. In general $\frac{d ^{n} x}{d t ^{n}} \leftrightarrow s^{n} X (s) - \sum_{k = 1}^{n} s^{n - k} x^{(k - 1)} (0^{-})$ . The values are taken at $0^{-}$ , before the input is applied.

UNILATERAL INTEGRATION

\int_{0^{-}}^{t} x (τ) d τ \leftrightarrow \frac{X ( s )}{s}, \int_{- \infty}^{t} x (τ) d τ \leftrightarrow \frac{X ( s )}{s} + \frac{1}{s} \int_{- \infty}^{0^{-}} x (τ) d τ

Unilateral transform. The second form carries the value the integral already had at $0^{-}$ , such as an initial capacitor voltage.

UNILATERAL TIME SHIFT (DELAY)

x (t - t_{0}) u (t - t_{0}) \leftrightarrow e^{- s t_{0}} X (s), t_{0} > 0

$X (s)$ is the transform of $x (t) u (t)$ . The whole causal signal must be delayed; $x (t - t_{0}) u (t)$ is a different signal and does not simply pick up $e^{- s t_{0}}$ .

DIVISION BY T (UNILATERAL)

\frac{x ( t )}{t} \leftrightarrow \int_{s}^{\infty} X (p) d p

Valid when $lim_{t \to 0^{+}} x (t) / t$ exists. Example: $\frac{sin t}{t} u (t) \leftrightarrow \int_{s}^{\infty} \frac{d p}{p ^{2} + 1} = \frac{π}{2} - tan^{- 1} s = tan^{- 1} \frac{1}{s}$ for real $s > 0$ .

WORKED EXAMPLE · CHAINING PROPERTIES

Find the Laplace transform, with ROC, of $x (t) = t e^{- 2 t} sin (3 t) u (t)$ .

Solution

Start from $sin (3 t) u (t) \leftrightarrow \frac{3}{s ^{2} + 9}$ , ROC $Re (s) > 0$ .
Shift in $s$ with $s_{0} = - 2$ : $e^{- 2 t} sin (3 t) u (t) \leftrightarrow \frac{3}{( s + 2 ) ^{2} + 9}$ , and the ROC moves to $Re (s) > - 2$ .
Multiply by $t$ : $X (s) = - \frac{d}{d s} [\frac{3}{( s + 2 ) ^{2} + 9}] = \frac{6 ( s + 2 )}{[ ( s + 2 ) ^{2} + 9 ] ^{2}}$ . The ROC is unchanged.
Sanity check: $s X (s) \to 0$ as $s \to \infty$ (it behaves like $6/ s^{2}$ ), matching $x (0^{+}) = 0$ .

Answer: $X (s) = \frac{6 ( s + 2 )}{( s ^{2} + 4 s + 13 ) ^{2}}$ , ROC $Re (s) > - 2$ .

EXAM TIP

Rewrite before you shift. For $x (t) = e^{- 2 t} u (t - 1)$ , write $x (t) = e^{- 2} e^{- 2 (t - 1)} u (t - 1)$ first, so $X (s) = \frac{e ^{- 2} e ^{- s}}{s + 2} = \frac{e ^{- (s + 2)}}{s + 2}$ with ROC $Re (s) > - 2$ . Dropping the constant $e^{- 2}$ is the usual slip.

EXAM TIP

Differentiation can enlarge the ROC. $u (t)$ has ROC $Re (s) > 0$ , but $\frac{d}{d t} u (t) = δ (t) \leftrightarrow s \cdot \frac{1}{s} = 1$ converges everywhere. That is why the table says the ROC contains $R$ rather than equals it.

Initial and final value theorems

Both theorems read a time-domain limit straight from $X (s)$ , with no inversion. Both are stated for signals that are zero for $t < 0$ , and both carry conditions that must be checked before the limit is trusted.

INITIAL VALUE THEOREM

x (0^{+}) = s \to \infty lim s X (s)

Valid when $x (t) = 0$ for $t < 0$ and $x (t)$ has no impulse (or higher singularity) at $t = 0$ ; for a rational $X (s)$ this means $X (s)$ is strictly proper (numerator degree less than denominator degree). If the numerator degree is equal or higher, divide first: the polynomial part is the impulse terms at the origin, which do not affect $x (0^{+})$ , and the theorem applies to the strictly proper remainder.

INITIAL SLOPE

x^{'} (0^{+}) = s \to \infty lim s [s X (s) - x (0^{+})]

Same conditions ( $X (s)$ strictly proper). Subtracting $x (0^{+})$ removes the impulse $x (0^{+}) δ (t)$ that a jump at $t = 0$ puts into the derivative.

FINAL VALUE THEOREM

t \to \infty lim x (t) = s \to 0 lim s X (s)

Valid, for a rational $X (s)$ , only if every pole of $s X (s)$ lies in the open LHP; equivalently, every pole of $X (s)$ is in the open LHP except at most a single simple pole at $s = 0$ . With a pole in the RHP, a repeated pole at the origin, or any other pole on the $j ω$ axis, $x (t)$ has no finite limit and the value the formula returns is not the final value.

$X (s)$	$lim_{s \to 0} s X (s)$	Theorem valid?	Actual $x (t)$
$\frac{4}{s ( s + 2 )}$	$2$	Yes: $s X (s)$ has its only pole at $- 2$	$2 (1 - e^{- 2 t}) \to 2$
$\frac{4}{s ( s - 2 )}$	$- 2$	No: pole at $+ 2$	$2 (e^{2 t} - 1)$ , grows without bound
$\frac{5}{s ^{2} + 25}$	$0$	No: poles at $\pm j 5$	$sin 5 t$ , oscillates with no limit
$\frac{1}{s ^{2}}$	$\infty$	No: double pole at $0$	$t$ , grows without bound

WORKED EXAMPLE · BOTH THEOREMS, THEN A CHECK BY INVERSION

A causal signal has $X (s) = \frac{2 s ^{2} + 7 s + 4}{s ( s + 1 ) ( s + 4 )}$ . Find $x (0^{+})$ and $lim_{t \to \infty} x (t)$ , and confirm both by finding $x (t)$ .

Solution

$X (s)$ is strictly proper (degree 2 over degree 3), so the IVT applies: $s X (s) = \frac{2 s ^{2} + 7 s + 4}{( s + 1 ) ( s + 4 )} \to 2$ as $s \to \infty$ . So $x (0^{+}) = 2$ .
The poles of $s X (s)$ are $- 1$ and $- 4$ , both in the open LHP, so the FVT applies: $lim_{s \to 0} s X (s) = \frac{4}{1 \cdot 4} = 1$ .
Check: $X (s) = \frac{A}{s} + \frac{B}{s + 1} + \frac{C}{s + 4}$ with $A = \frac{4}{4} = 1$ , $B = \frac{2 - 7 + 4}{( - 1 ) ( 3 )} = \frac{1}{3}$ , $C = \frac{32 - 28 + 4}{( - 4 ) ( - 3 )} = \frac{2}{3}$ .
$x (t) = (1 + \frac{1}{3} e^{- t} + \frac{2}{3} e^{- 4 t}) u (t)$ , so $x (0^{+}) = 1 + \frac{1}{3} + \frac{2}{3} = 2$ and $x (\infty) = 1$ .

Answer: $x (0^{+}) = 2$ , $lim_{t \to \infty} x (t) = 1$ .

EXAM TIP

Divide first when the degrees are equal. For $X (s) = \frac{s + 3}{s + 1}$ , $lim_{s \to \infty} s X (s)$ is infinite, but $X (s) = 1 + \frac{2}{s + 1}$ means $x (t) = δ (t) + 2 e^{- t} u (t)$ , so $x (0^{+}) = 2$ . Apply the IVT to $\frac{2}{s + 1}$ only.

EXAM TIP

Check the poles of $s X (s)$ , not of $X (s)$ . $X (s) = \frac{1}{s - 1}$ gives $lim_{s \to 0} s X (s) = 0$ , yet $x (t) = e^{t} u (t)$ diverges. For the unit-step response of a causal stable $H (s)$ , $Y (s) = H (s) / s$ and $s Y (s) = H (s)$ , so the steady-state value is simply $H (0)$ , the DC gain.

Inverse transform by partial fractions

For a rational $X (s) = N (s) / D (s)$ , first make it strictly proper: if $de g N \geq de g D$ , divide, so that $X (s)$ is a polynomial in $s$ plus a strictly proper remainder. The polynomial terms invert to $δ (t), δ^{'} (t), \dots$ . Then expand the remainder over its poles and invert term by term, choosing a right-sided or left-sided pair for each pole according to where it lies relative to the ROC.

Procedure

Divide if needed: a constant $c$ inverts to $c δ (t)$ and a term $cs$ to $c δ^{'} (t)$ .
Factor the denominator. Write one term per simple pole, $r$ terms for a pole of order $r$ , and one quadratic term (or a conjugate pair of terms) for each complex pair.
Find the coefficients with the residue formulas below, then check by substituting one convenient value such as $s = 0$ into both sides.
For each term: a pole to the left of the ROC gives a right-sided term with $u (t)$ ; a pole to the right of the ROC gives a left-sided term with $- u (- t)$ .
Add the terms. In a causal (unilateral) problem every term is right-sided.

COEFFICIENT AT A SIMPLE POLE (COVER-UP)

X (s) = k \sum \frac{A _{k}}{s - p _{k}}, A_{k} = (s - p_{k}) X (s)_{s = p_{k}}

Each term inverts to $A_{k} e^{p_{k} t} u (t)$ if the ROC is to the right of $p_{k}$ , or to $- A_{k} e^{p_{k} t} u (- t)$ if the ROC is to the left of $p_{k}$ .

COEFFICIENTS AT A POLE OF ORDER R

m = 1 \sum r \frac{B _{m}}{( s - p ) ^{m}}, B_{m} = \frac{1}{( r - m )!} \frac{d ^{r - m}}{d s ^{r - m}} [(s - p)^{r} X (s)]_{s = p}

Only the top coefficient $B_{r}$ is plain cover-up; each lower one needs one more derivative. Inversion: $\frac{1}{( s - p ) ^{m}} \leftrightarrow \frac{t ^{m - 1}}{( m - 1 )!} e^{pt} u (t)$ for an ROC to the right of $p$ , and $- \frac{t ^{m - 1}}{( m - 1 )!} e^{pt} u (- t)$ for an ROC to the left.

COMPLEX-CONJUGATE POLE PAIR (COMPLETE THE SQUARE)

\frac{A s + B}{( s + a ) ^{2} + b ^{2}} = A \frac{s + a}{( s + a ) ^{2} + b ^{2}} + \frac{B - A a}{b} \cdot \frac{b}{( s + a ) ^{2} + b ^{2}} \leftrightarrow e^{- a t} [A cos b t + \frac{B - A a}{b} sin b t] u (t)

Causal case, ROC $Re (s) > - a$ , with $A$ , $B$ , $a$ , $b$ real and $b > 0$ . Residue form of the same result, for a real signal: if the pole $p = - a + j b$ has residue $K$ , the pair contributes $2∣ K ∣ e^{- a t} cos (b t + ∠ K) u (t)$ .

WORKED EXAMPLE · REPEATED POLE, TWO DIFFERENT ROCS

Find $x (t)$ for $X (s) = \frac{s + 3}{( s + 1 ) ^{2} ( s + 2 )}$ when (a) the ROC is $Re (s) > - 1$ and (b) the ROC is $- 2 < Re (s) < - 1$ .

Solution

Form: $X (s) = \frac{A}{s + 2} + \frac{B _{1}}{s + 1} + \frac{B _{2}}{( s + 1 ) ^{2}}$ .
$A = \frac{s + 3}{( s + 1 ) ^{2}}_{s = - 2} = \frac{1}{1} = 1$ and $B_{2} = \frac{s + 3}{s + 2}_{s = - 1} = 2$ .
$B_{1} = \frac{d}{d s} [\frac{s + 3}{s + 2}]_{s = - 1} = \frac{- 1}{( s + 2 ) ^{2}}_{s = - 1} = - 1$ .
Check at $s = 0$ : $X (0) = \frac{3}{2}$ and $\frac{1}{2} - 1 + 2 = \frac{3}{2}$ .
(a) Both poles are left of the ROC: $x (t) = (e^{- 2 t} - e^{- t} + 2 t e^{- t}) u (t)$ .
(b) The pole at $- 2$ is left of the ROC (right-sided term); the pole at $- 1$ is right of it (left-sided terms, each with a minus sign): $x (t) = e^{- 2 t} u (t) + (e^{- t} - 2 t e^{- t}) u (- t)$ .

Answer: (a) $(e^{- 2 t} - e^{- t} + 2 t e^{- t}) u (t)$ ; (b) $e^{- 2 t} u (t) + (e^{- t} - 2 t e^{- t}) u (- t)$ .

WORKED EXAMPLE · COMPLEX POLES

Find the causal $x (t)$ whose transform is $X (s) = \frac{2 s + 8}{s ^{2} + 2 s + 5}$ .

Solution

Complete the square: $s^{2} + 2 s + 5 = (s + 1)^{2} + 2^{2}$ , so $a = 1$ , $b = 2$ , and the numerator gives $A = 2$ , $B = 8$ .
$X (s) = 2 \frac{s + 1}{( s + 1 ) ^{2} + 4} + \frac{8 - 2}{2} \cdot \frac{2}{( s + 1 ) ^{2} + 4}$ .
$x (t) = e^{- t} (2 cos 2 t + 3 sin 2 t) u (t)$ .
Residue check: at $p = - 1 + j 2$ , $K = \frac{2 p + 8}{p - p ^{*}} = \frac{6 + j 4}{j 4} = 1 - j 1.5$ , so $2∣ K ∣ = 13$ and $∠ K = - 56.3 1^{\circ}$ , giving $13 e^{- t} cos (2 t - 56.3 1^{\circ}) u (t)$ , the same signal.
IVT check: $s X (s) \to 2$ as $s \to \infty$ , and $x (0^{+}) = 2$ .

Answer: $x (t) = e^{- t} (2 cos 2 t + 3 sin 2 t) u (t) = 13 e^{- t} cos (2 t - 56.3 1^{\circ}) u (t)$ .

EXAM TIP

Cover-up gives only the top coefficient of a repeated pole. Covering $(s + 1)^{2}$ in the example gives $B_{2}$ ; using the same trick for $B_{1}$ gives a wrong number. Differentiate, or substitute a spare value of $s$ once the other coefficients are known.

EXAM TIP

An improper $X (s)$ hides impulses. $X (s) = \frac{s ^{2} + 3 s + 5}{s + 1} = s + 2 + \frac{3}{s + 1}$ , so the causal inverse is $δ^{'} (t) + 2 δ (t) + 3 e^{- t} u (t)$ . Expanding without dividing first loses the $δ^{'} (t)$ and $δ (t)$ terms.

System function: poles, zeros, causality and stability

For an LTI system with impulse response $h (t)$ , the convolution property turns $y (t) = x (t) * h (t)$ into $Y (s) = H (s) X (s)$ . The system function (transfer function) $H (s) = L {h (t)}$ is the ratio $Y (s) / X (s)$ with every initial condition zero. Its ROC is part of the system: one rational $H (s)$ can describe a causal, an anti-causal or a two-sided system.

The exponential $e^{s_{0} t}$ is an eigenfunction: an input $e^{s_{0} t}$ applied for all $t$ , with $s_{0}$ in the ROC, produces the output $H (s_{0}) e^{s_{0} t}$ .

TRANSFER FUNCTION FROM A DIFFERENTIAL EQUATION

k = 0 \sum N a_{k} \frac{d ^{k} y}{d t ^{k}} = k = 0 \sum M b_{k} \frac{d ^{k} x}{d t ^{k}} ⟹ H (s) = \frac{Y ( s )}{X ( s )} = \frac{\sum _{k = 0}^{M} b _{k} s ^{k}}{\sum _{k = 0}^{N} a _{k} s ^{k}}

Replace each $\frac{d ^{k}}{d t ^{k}}$ by $s^{k}$ , with zero initial conditions. The equation alone does not fix the ROC; a statement such as causal, stable or initially at rest does.

POLE-ZERO-GAIN FORM

H (s) = K \frac{( s - z _{1} ) ( s - z _{2} ) \dots ( s - z _{M} )}{( s - p _{1} ) ( s - p _{2} ) \dots ( s - p _{N} )}

Zeros $z_{i}$ are the roots of the numerator and poles $p_{k}$ the roots of the denominator, after cancelling common factors. $H (s)$ is proper if $M \leq N$ and strictly proper if $M < N$ . On a pole-zero plot poles are marked $\times$ and zeros $\circ$ . For real $h (t)$ the coefficients are real, so complex poles and zeros come in conjugate pairs. The plot fixes $H (s)$ only up to the gain $K$ .

In a causal system each pole $p_{k}$ contributes a mode $e^{p_{k} t}$ to $h (t)$ . A real LHP pole gives a decaying exponential, a complex pair $σ \pm j ω_{d}$ gives $e^{σ t} cos (ω_{d} t + ϕ)$ , a simple pole on the $j ω$ axis gives a constant (pole at $0$ ) or a sustained sinusoid (pair at $\pm j ω_{0}$ ), and a repeated pole multiplies its mode by powers of $t$ . Zeros create no modes: they set how much of each mode appears (the residues), and an everlasting input $e^{z_{i} t}$ with $z_{i}$ in the ROC gives zero output.

POLES OF A SECOND-ORDER DENOMINATOR

s^{2} + 2 ζ ω_{n} s + ω_{n}^{2} = 0 ⟹ p_{1, 2} = - ζ ω_{n} \pm j ω_{n} 1 - ζ^{2} (ω_{n} > 0, 0 \leq ζ < 1)

For $0 < ζ < 1$ the poles lie on a circle of radius $ω_{n}$ , at angle $cos^{- 1} ζ$ from the negative real axis; $ζ = 0$ puts them on the $j ω$ axis at $\pm j ω_{n}$ . For $ζ = 1$ there is a double pole at $- ω_{n}$ , and for $ζ > 1$ two real poles $- ζ ω_{n} \pm ω_{n} ζ^{2} - 1$ . A negative $ζ$ puts the poles in the RHP.

INTERCONNECTED SYSTEMS

cascade: H_{1} (s) H_{2} (s), parallel: H_{1} (s) + H_{2} (s), negative feedback: \frac{G ( s )}{1 + G ( s ) F ( s )}

The cascade result assumes the second block does not load the first. In the feedback loop $G (s)$ is the forward path and $F (s)$ the feedback path; with positive feedback the denominator is $1 - G (s) F (s)$ . Writing $G = N_{G} / D_{G}$ and $F = N_{F} / D_{F}$ , the negative-feedback system is $\frac{N _{G} D _{F}}{D _{G} D _{F} + N _{G} N _{F}}$ , so its poles are the roots of $D_{G} D_{F} + N_{G} N_{F}$ (after cancelling any common factor), not the poles of $G$ or $F$ .

System property	Condition on the ROC	For rational $H (s)$ , in terms of poles
Causal: $h (t) = 0$ for $t < 0$	a right half-plane (necessary; sufficient only for rational $H (s)$ )	ROC to the right of the rightmost pole
Anti-causal: $h (t) = 0$ for $t > 0$	a left half-plane (necessary; sufficient only for rational $H (s)$ )	ROC to the left of the leftmost pole
BIBO stable: $\int_{- \infty}^{\infty} ∣ h (t) ∣ d t < \infty$	contains the $j ω$ axis	$H (s)$ proper, no pole on the $j ω$ axis, and the ROC strip includes $Re (s) = 0$
Causal and stable	a right half-plane that contains the $j ω$ axis	every pole in the open LHP, $Re (p_{k}) < 0$ , with $H (s)$ proper

WORKED EXAMPLE · FROM A DIFFERENTIAL EQUATION TO H(T)

A causal LTI system obeys $\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + 2 y = \frac{d x}{d t} + 3 x$ . Find $H (s)$ with its poles, zeros and ROC, the impulse response $h (t)$ , and the final value of the unit-step response.

Solution

With zero initial conditions, $(s^{2} + 3 s + 2) Y (s) = (s + 3) X (s)$ , so $H (s) = \frac{s + 3}{( s + 1 ) ( s + 2 )}$ .
Zero at $s = - 3$ ; poles at $s = - 1$ and $s = - 2$ . Causal means the ROC is $Re (s) > - 1$ ; it contains the $j ω$ axis, so the system is also stable.
Cover-up: $\frac{- 1 + 3}{- 1 + 2} = 2$ and $\frac{- 2 + 3}{- 2 + 1} = - 1$ , so $H (s) = \frac{2}{s + 1} - \frac{1}{s + 2}$ and $h (t) = (2 e^{- t} - e^{- 2 t}) u (t)$ .
Step response: $Y (s) = H (s) / s$ , and $s Y (s) = H (s)$ has all its poles in the LHP, so $y (\infty) = H (0) = \frac{3}{2}$ .

Answer: $H (s) = \frac{s + 3}{( s + 1 ) ( s + 2 )}$ , ROC $Re (s) > - 1$ ; zero at $- 3$ , poles at $- 1, - 2$ ; $h (t) = (2 e^{- t} - e^{- 2 t}) u (t)$ ; the step response settles at $1.5$ .

EXAM TIP

A pole in the RHP does not by itself mean unstable. $H (s) = \frac{5}{( s + 2 ) ( s - 3 )} = - \frac{1}{s + 2} + \frac{1}{s - 3}$ with ROC $- 2 < Re (s) < 3$ contains the $j ω$ axis, so it is stable, with $h (t) = - e^{- 2 t} u (t) - e^{3 t} u (- t)$ . It is not causal. Only a system that is both causal and stable must have every pole in the LHP.

EXAM TIP

Poles on the $j ω$ axis mean not BIBO stable. The causal integrator $H (s) = \frac{1}{s}$ turns a bounded step into a ramp. The causal $H (s) = \frac{1}{s ^{2} + 4}$ , initially at rest and driven by $sin (2 t) u (t)$ , gives $Y (s) = \frac{2}{( s ^{2} + 4 ) ^{2}}$ , so $y (t) = (\frac{sin 2 t}{8} - \frac{t cos 2 t}{4}) u (t)$ , which grows with $t$ . A causal system with proper $H (s)$ , simple poles on the $j ω$ axis and every other pole in the LHP is often called marginally stable: its $h (t)$ stays bounded, but it is still not BIBO stable. An improper $H (s)$ is not BIBO stable either: the differentiator $H (s) = s$ has $h (t) = δ^{'} (t)$ .

EXAM TIP

Stability of a low-order causal system from its coefficients. Take the denominator of $H (s)$ after cancelling common factors, with a positive leading coefficient. For first order ( $s + a_{0}$ ) or second order ( $s^{2} + a_{1} s + a_{0}$ ), every root is in the open LHP if and only if every coefficient is positive. From third order on, positive coefficients are necessary but not sufficient: $s^{3} + s^{2} + s + 1 = (s + 1) (s^{2} + 1)$ has roots at $\pm j$ . At any order, a zero or negative coefficient means some root is not in the open LHP.

EXAM TIP

Cancel common factors, but know what the cancellation hides. $H (s) = \frac{s - 1}{( s - 1 ) ( s + 2 )} = \frac{1}{s + 2}$ is the input-output description, and its causal version is BIBO stable. A realisation that still contains the factor $(s - 1)$ has an internal mode $e^{t}$ that the input cannot excite and the output cannot see, but a non-zero initial state can; such a system is not internally stable.

EXAM TIP

For a non-rational $H (s)$ , a right-half-plane ROC does not prove causality. $H (s) = \frac{e ^{s}}{s + 1}$ with ROC $Re (s) > - 1$ is $h (t) = e^{- (t + 1)} u (t + 1)$ , which is non-zero for $- 1 < t < 0$ . The pole test in the table is for rational $H (s)$ .

Frequency response, all-pass and minimum-phase systems

For a stable system the ROC contains the $j ω$ axis, so the frequency response is $H (s)$ evaluated on that axis. For real $h (t)$ , $H (- j ω) = H^{*} (j ω)$ : $∣ H (j ω) ∣$ is even and $∠ H (j ω)$ is odd in $ω$ , and a sinusoid comes out as a sinusoid of the same frequency, scaled and phase-shifted.

SINUSOIDAL STEADY STATE

x (t) = A cos (ω_{0} t + θ) ⟹ y (t) = A H (j ω_{0}) cos (ω_{0} t + θ + ∠ H (j ω_{0}))

Stable LTI system with real $h (t)$ ; the input is applied for all time, or $y (t)$ is the steady state after the transients have died out. $H (j ω) = H (s)_{s = j ω}$ .

GEOMETRIC (VECTOR) EVALUATION

H (j ω) = ∣ K ∣ \frac{\prod _{i} j ω - z _{i}}{\prod _{k} j ω - p _{k}}, ∠ H (j ω) = ∠ K + i \sum ∠ (j ω - z_{i}) - k \sum ∠ (j ω - p_{k})

$j ω - z_{i}$ is the vector from the zero $z_{i}$ to the point $j ω$ on the axis; likewise for each pole. Magnitude: product of zero-vector lengths over product of pole-vector lengths. Phase: sum of zero-vector angles minus sum of pole-vector angles, each measured anticlockwise from the positive real direction. $∠ K = 18 0^{\circ}$ if $K < 0$ .

Reading a pole-zero plot

A pole pair close to the $j ω$ axis at $- σ \pm j ω_{0}$ (small $σ > 0$ ) makes one pole vector short for $ω$ near $ω_{0}$ , giving a sharp peak (resonance) near $ω_{0}$ .
A zero on the $j ω$ axis at $\pm j ω_{0}$ makes $∣ H (j ω_{0}) ∣ = 0$ : that frequency is blocked (a notch).
A zero at the origin makes $∣ H (j 0) ∣ = 0$ (no DC gain). With more poles than zeros ( $N > M$ ), $∣ H (j ω) ∣ \to 0$ as $ω \to \infty$ , falling like $ω^{- (N - M)}$ .
As $ω$ runs from $0$ to $\infty$ , the net phase change is $- 9 0^{\circ}$ for each real LHP pole, $+ 9 0^{\circ}$ for each real LHP zero, $- 9 0^{\circ}$ for each real RHP zero, $- 18 0^{\circ}$ for each complex LHP pole pair, $+ 18 0^{\circ}$ for each complex LHP zero pair and $- 18 0^{\circ}$ for each complex RHP zero pair. A pole or zero at the origin adds a constant $- 9 0^{\circ}$ or $+ 9 0^{\circ}$ for $ω > 0$ .

WORKED EXAMPLE · VECTOR EVALUATION AT ONE FREQUENCY

A causal system has $H (s) = \frac{10 ( s + 1 )}{s ^{2} + 2 s + 5}$ . Find $∣ H (j 2) ∣$ and $∠ H (j 2)$ , and the steady-state output for the input $cos 2 t$ .

Solution

Zero at $- 1$ ; poles at $- 1 \pm j 2$ (roots of $s^{2} + 2 s + 5$ ), both in the LHP, so the system is stable; $K = 10$ .
Vectors to $j 2$ . From the zero: $j 2 - (- 1) = 1 + j 2$ , length $5$ , angle $tan^{- 1} 2 = 63.4 3^{\circ}$ . From $- 1 + j 2$ : $1 + j 0$ , length $1$ , angle $0^{\circ}$ . From $- 1 - j 2$ : $1 + j 4$ , length $17$ , angle $tan^{- 1} 4 = 75.9 6^{\circ}$ .
$∣ H (j 2) ∣ = \frac{10 5}{1 \cdot 17} = 5.42$ and $∠ H (j 2) = 63.4 3^{\circ} - 0^{\circ} - 75.9 6^{\circ} = - 12.5 3^{\circ}$ .
Direct check: $H (j 2) = \frac{10 ( 1 + j 2 )}{- 4 + j 4 + 5} = \frac{10 ( 1 + j 2 )}{1 + j 4} = \frac{10 ( 9 - j 2 )}{17}$ , with magnitude $\frac{10 85}{17} = 5.42$ and angle $- tan^{- 1} \frac{2}{9} = - 12.5 3^{\circ}$ .
The pole at $- 1 + j 2$ is only one unit from $j 2$ , which is why the gain is large near $ω = 2$ rad/s.

Answer: $∣ H (j 2) ∣ \approx 5.42$ , $∠ H (j 2) \approx - 12.5 3^{\circ}$ ; the output is $5.42 cos (2 t - 12.5 3^{\circ})$ .

An all-pass system has $∣ H (j ω) ∣$ constant for every $ω$ . For a causal stable rational all-pass, each LHP pole $p$ is paired with a zero at its mirror image $- p^{*}$ in the RHP. Each zero vector then has the same length as its partner pole vector at every $ω$ , so only the phase changes.

A minimum-phase system is causal and stable with every zero in the open LHP as well. Then $1/ H (s)$ also has every pole in the open LHP; its causal version is stable provided it is proper, which needs the numerator and denominator of $H (s)$ to have equal degree. Any causal stable rational $H (s)$ with no zeros on the $j ω$ axis factors as $H (s) = H_{m i n} (s) H_{ap} (s)$ : reflect each RHP zero into the LHP to get $H_{m i n} (s)$ , and the all-pass factor carries the reflection.

FIRST-ORDER ALL-PASS

H_{ap} (s) = \frac{a - s}{a + s}, a > 0 : H_{ap} (j ω) = 1, ∠ H_{ap} (j ω) = - 2 tan^{- 1} \frac{ω}{a}

Pole at $- a$ , zero at $+ a$ . The phase falls from $0$ to $- 18 0^{\circ}$ ; the group delay is $τ (ω) = - \frac{d ∠ H}{d ω} = \frac{2 a}{a ^{2} + ω ^{2}}$ . The form $\frac{s - a}{s + a}$ is the same system times $- 1$ , with phase $18 0^{\circ} - 2 tan^{- 1} (ω / a)$ .

MINIMUM-PHASE CONDITION

H (s) is minimum phase ⟺ causal, stable, all poles and all zeros in Re (s) < 0

Among causal stable systems with the same $∣ H (j ω) ∣$ , the minimum-phase one has the least group delay at every $ω$ and the least net phase lag over $0 \leq ω < \infty$ , because each all-pass factor has positive group delay and only adds lag. When the numerator and denominator degrees are equal, $1/ H (s)$ is also causal and stable. Some texts also allow zeros on the $j ω$ axis; the open-LHP version is the one used here.

EXAM TIP

Equal magnitude responses, different phase. $\frac{s + 2}{s + 5}$ and $\frac{2 - s}{s + 5}$ have identical $∣ H (j ω) ∣$ , because the vectors from $- 2$ and from $+ 2$ to any point $j ω$ have the same length. The first is minimum phase; the second equals $\frac{s + 2}{s + 5} \cdot \frac{2 - s}{2 + s}$ , minimum phase times all-pass. A question that gives only $∣ H (j ω) ∣$ cannot tell them apart.

EXAM TIP

No $j ω$ -axis ROC, no frequency response. For $u (t)$ , $\frac{1}{s}_{s = j ω} = \frac{1}{j ω}$ is not its Fourier transform; the correct one is $π δ (ω) + \frac{1}{j ω}$ . Substitute $s = j ω$ only when the ROC contains the axis.

Solving LTI differential equations with initial conditions

The unilateral transform turns a linear constant-coefficient differential equation with initial conditions into an algebraic equation for $Y (s)$ , because the differentiation property brings the initial values in automatically. The solution splits into a part driven by the initial conditions (the zero-input response) and a part driven by the input (the zero-state response).

ZERO-INPUT PLUS ZERO-STATE

Y (s) = zero-input \frac{I ( s )}{A ( s )} + zero-state H (s) X (s), H (s) = \frac{B ( s )}{A ( s )}

$A (s)$ and $B (s)$ are the polynomials that multiply $Y (s)$ and $X (s)$ . $I (s)$ collects the initial-condition terms $y (0^{-}), y^{'} (0^{-}), \dots$ (terms in $x (0^{-})$ vanish for a causal input). The natural modes come from the roots of $A (s)$ , the forced part from the poles of $X (s)$ .

Procedure

Transform both sides with the unilateral rules $y^{'} \to s Y - y (0^{-})$ and $y^{''} \to s^{2} Y - s y (0^{-}) - y^{'} (0^{-})$ .
Collect terms: $A (s) Y (s) = I (s) + B (s) X (s)$ , and solve for $Y (s)$ .
Expand in partial fractions, keeping the zero-input and zero-state parts separate if they are asked for, and invert with right-sided pairs. The result holds for $t \geq 0$ .
Check $y (0^{+})$ with the IVT and $y (\infty)$ with the FVT (when valid) against the answer.

WORKED EXAMPLE · SECOND-ORDER EQUATION WITH INITIAL CONDITIONS

Solve $\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + 2 y = 2 u (t)$ with $y (0^{-}) = 1$ and $y^{'} (0^{-}) = - 1$ . Identify the zero-input and zero-state responses.

Solution

Transform: $s^{2} Y - s (1) - (- 1) + 3 [s Y - 1] + 2 Y = \frac{2}{s}$ .
Collect: $(s^{2} + 3 s + 2) Y = s + 2 + \frac{2}{s}$ , so $Y (s) = \frac{s + 2}{( s + 1 ) ( s + 2 )} + \frac{2}{s ( s + 1 ) ( s + 2 )}$ .
Zero-input part: $\frac{s + 2}{( s + 1 ) ( s + 2 )} = \frac{1}{s + 1}$ , giving $e^{- t}$ . These initial conditions excite only the $e^{- t}$ mode.
Zero-state part: $\frac{2}{s ( s + 1 ) ( s + 2 )} = \frac{1}{s} - \frac{2}{s + 1} + \frac{1}{s + 2}$ , giving $1 - 2 e^{- t} + e^{- 2 t}$ .
Total: $y (t) = 1 - e^{- t} + e^{- 2 t}$ for $t \geq 0$ . Check: $y (0) = 1$ ; $y^{'} (t) = e^{- t} - 2 e^{- 2 t}$ , so $y^{'} (0) = - 1$ ; and $y (\infty) = 1 = lim_{s \to 0} s Y (s)$ .

Answer: $y (t) = 1 - e^{- t} + e^{- 2 t}$ for $t \geq 0$ ; zero-input $e^{- t}$ , zero-state $1 - 2 e^{- t} + e^{- 2 t}$ .

CIRCUIT ELEMENTS IN THE S-DOMAIN

V_{R} (s) = R I_{R} (s), V_{L} (s) = s L I_{L} (s) - L i_{L} (0^{-}), V_{C} (s) = \frac{I _{C} ( s )}{s C} + \frac{v _{C} ( 0 ^{-} )}{s}

Unilateral transform of $v = R i$ , $v = L \frac{d i}{d t}$ and $i = C \frac{d v}{d t}$ , with the passive sign convention (current entering the $+$ terminal). The terms $- L i_{L} (0^{-})$ and $\frac{v _{C} ( 0 ^{-} )}{s}$ act as voltage sources in series with the impedances $s L$ and $\frac{1}{s C}$ . With zero initial conditions only the impedances $R$ , $s L$ and $\frac{1}{s C}$ remain, and a circuit's transfer function follows by ordinary impedance algebra.

EXAM TIP

Use the $0^{-}$ values. The unilateral transform with lower limit $0^{-}$ needs the conditions just before the input is applied. If the right-hand side (after any differentiation of $x$ ) has no impulse at $t = 0$ , then $y, y^{'}, \dots, y^{(N - 1)}$ are continuous at $t = 0$ and the $0^{+}$ and $0^{-}$ values agree. If it does contain an impulse, some of them differ, and putting $0^{+}$ values into the $0^{-}$ formula is an error.

EXAM TIP

Zero-state is not forced, and zero-input is not natural. The zero-state response $1 - 2 e^{- t} + e^{- 2 t}$ above contains natural modes too. The natural response is every term from the roots of $A (s)$ , here $- e^{- t} + e^{- 2 t}$ ; the forced response is the term from the input pole, here $1$ .

EXAM TIP

Impulse, step and ramp responses. For a causal system at rest, $h (t)$ is the zero-state response to $δ (t)$ . The unit-step response is $\int_{0^{-}}^{t} h (τ) d τ \leftrightarrow \frac{H ( s )}{s}$ , so differentiating the step response gives $h (t)$ . The unit-ramp response, $\leftrightarrow \frac{H ( s )}{s ^{2}}$ , is the integral of the step response.

PREVIOUS-YEAR QUESTIONS

8 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE EE 2024 · GATE EE 2020 · GATE IN 2026 · GATE IN 2024 · GATE IN 2023 · GATE IN 2022

GATE 2026 EC Q.13

GATE 2026 · ECQ.13 · 1 mark · MCQ

The Laplace Transform of the signal $x (t) = u (t - 2) * (t u (t))$ is given by which of the following expressions?

[" $*$ " represents convolution operator]

(A)
$\frac{e ^{- 2 s}}{s ^{2} ( s - 2 )}$
(B)
$\frac{e ^{- 2 (s - 2)}}{s ^{3}}$
(C)
$\frac{s e ^{- 2 s}}{( s - 2 ) ^{2}}$
(D)
$\frac{e ^{- 2 s}}{s ^{3}}$

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GATE 2026 IN Q.43

GATE 2026 · INQ.43 · 2 marks · MCQ

Given two signals: $x (t) = u (t - 2) - u (t - 3)$ , and $y (t) = e^{- 4 t} u (t)$ .

If $z (t) = x (t) * y (t)$ , then $Z (s)$ is ________.

(A)
$\frac{e ^{- 3 s} [ e ^{s} - 1 ]}{s ( s + 4 )}$
(B)
$\frac{e ^{- s} [ e ^{- 3 s} - 1 ]}{( s + 3 ) ( s + 4 )}$
(C)
$\frac{e ^{- 3 s} [ 1 - e ^{s} ]}{( s + 3 ) ( s + 4 )}$
(D)
$\frac{e ^{- 3 s} [ 1 - e ^{s} ]}{s ( s + 4 )}$

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GATE 2024 EE Q.16

GATE 2024 · EEQ.16 · 1 mark · MCQ

If $u (t)$ is the unit step function, then the region of convergence (ROC) of the Laplace transform of the signal

x (t) = e^{t^{2}} [u (t - 1) - u (t - 10)]

(A)
$- \infty < Re (s) < \infty$
(B)
$Re (s) \geq 10$
(C)
$Re (s) \leq 1$
(D)
$1 \leq Re (s) \leq 10$

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GATE 2024 IN Q.40

GATE 2024 · INQ.40 · 2 marks · MCQ

Laplace transform of a signal $x (t)$ is

X (s) = \frac{1}{s ^{2} + 13 s + 42}

Let $u (t)$ be the unit step function. Choose the signal $x (t)$ from the following options if the region of convergence is $- 7 < Re {s} < - 6$ .

(A)
$- e^{- 6 t} u (t) - e^{- 7 t} u (- t)$
(B)
$- e^{- 6 t} u (- t) - e^{- 7 t} u (t)$
(C)
$e^{- 6 t} u (t) - e^{- 7 t} u (- t)$
(D)
$- e^{- 6 t} u (- t) - e^{- 7 t} u (- t)$

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GATE 2023 IN Q.37

GATE 2023 · INQ.37 · 2 marks · MCQ

The Laplace transform of the continuous-time signal $x (t) = e^{- 3 t} u (t - 5)$ is ______, where $u (t)$ denotes the continuous-time unit step signal.

(A)
$\frac{e ^{- 5 s}}{s + 3}, R e a l {s} > - 3$
(B)
$\frac{e ^{- 5 (s - 3)}}{s - 3}, R e a l {s} > 3$
(C)
$\frac{e ^{- 5 (s + 3)}}{s + 3}, R e a l {s} > - 3$
(D)
$\frac{e ^{- 5 (s - 3)}}{s + 3}, R e a l {s} > - 3$

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GATE 2022 IN Q.40

GATE 2022 · INQ.40 · 2 marks · MCQ

The signal $x (t) = (t - 1)^{2} u (t - 1)$ , where $u (t)$ is the unit-step function, has the Laplace transform $X (s)$ . The value of $X (1)$ is ________

(A)
$\frac{1}{e}$
(B)
$\frac{2}{e}$
(C)
$2 e$
(D)
$e^{2}$

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GATE 2020 EE Q.3

GATE 2020 · EEQ.3 · 1 mark · MCQ

Which of the following statements is true about the two sided Laplace transform?

(A)
It exists for every signal that may or may not have a Fourier transform.
(B)
It has no poles for any bounded signal that is non-zero only inside a finite time interval.
(C)
The number of finite poles and finite zeroes must be equal.
(D)
If a signal can be expressed as a weighted sum of shifted one sided exponentials, then its Laplace Transform will have no poles.

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GATE 2020 EE Q.38

GATE 2020 · EEQ.38 · 2 marks · MCQ

The causal realization of a system transfer function $H (s)$ having poles at $(2, - 1)$ , $(- 2, 1)$ and zeroes at $(2, 1)$ , $(- 2, - 1)$ will be

(A)
stable, real, allpass
(B)
unstable, complex, allpass
(C)
unstable, real, highpass
(D)
stable, complex, lowpass

Answer and solution

GATE 2020 EE Q.38: answer and solution

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GATE 2020 EE Q.38: answer and solution

PRACTICE

7 questions to test yourself

PRACTICE 11 mark · NAT

A causal signal $x (t)$ has Laplace transform $X (s) = \frac{5 ( s + 4 )}{s ( s ^{2} + 3 s + 10 )}$ . The value of $lim_{t \to \infty} x (t)$ is ______.

Answer and solution

Answer: $2$ (accepted range $\pm 0.01$ )

$s X (s) = \frac{5 ( s + 4 )}{s ^{2} + 3 s + 10}$ . Its poles are the roots of $s^{2} + 3 s + 10$ , namely $- 1.5 \pm j 2.78$ , both in the open LHP, so the final value theorem applies.
$lim_{s \to 0} s X (s) = \frac{5 \cdot 4}{10} = 2$ .

PRACTICE 21 mark · MCQ

For $x (t) = e^{- 2 t} u (- t) + e^{- 4 t} u (t)$ , which statement about its Laplace transform is correct?

(A)
The ROC is $- 4 < Re (s) < - 2$ , and the Fourier transform of $x (t)$ exists
(B)
The ROC is $- 4 < Re (s) < - 2$ , and the Fourier transform of $x (t)$ does not exist
(C)
The ROC is $Re (s) > - 2$ , and the Fourier transform of $x (t)$ exists
(D)
The ROC is empty, so $x (t)$ has no Laplace transform

Answer and solution

Answer: (B)

$e^{- 4 t} u (t)$ is right-sided and converges for $Re (s) > - 4$ .
$e^{- 2 t} u (- t)$ is left-sided: $\int_{- \infty}^{0} e^{- (s + 2) t} d t$ converges for $Re (s) < - 2$ .
The intersection $- 4 < Re (s) < - 2$ is non-empty, so $X (s) = \frac{1}{s + 4} - \frac{1}{s + 2}$ exists there.
The strip does not contain $Re (s) = 0$ . Indeed $e^{- 2 t} u (- t)$ grows without bound as $t \to - \infty$ , so $x (t)$ has no Fourier transform.

PRACTICE 31 mark · MCQ

A causal LTI system has $H (s) = \frac{s - 2}{s + 2}$ . The input $x (t) = cos (2 t)$ is applied for all $t$ . The output $y (t)$ is

(A)
$sin (2 t)$
(B)
$- sin (2 t)$
(C)
$cos (2 t)$
(D)
$\frac{1}{2} cos (2 t - 4 5^{\circ})$

Answer and solution

Answer: (B)

The only pole is at $- 2$ , so the causal system is stable and the sinusoidal steady-state formula applies; with the input applied for all $t$ there is no transient.
$H (j 2) = \frac{- 2 + j 2}{2 + j 2}$ . The numerator has length $22$ and angle $13 5^{\circ}$ ; the denominator has length $22$ and angle $4 5^{\circ}$ .
So $∣ H (j 2) ∣ = 1$ and $∠ H (j 2) = 13 5^{\circ} - 4 5^{\circ} = 9 0^{\circ}$ (indeed $H (j 2) = j$ ); the system is all-pass.
$y (t) = cos (2 t + 9 0^{\circ}) = - sin (2 t)$ .

PRACTICE 41 mark · NAT

A causal signal has Laplace transform $X (s) = \frac{2 s ^{2} + 5 s + 1}{s ^{2} + 3 s + 2}$ . The value of $x (0^{+})$ is ______.

Answer and solution

Answer: $- 1$ (accepted range $\pm 0.01$ )

$X (s)$ is not strictly proper, so divide first: $2 s^{2} + 5 s + 1 = 2 (s^{2} + 3 s + 2) - (s + 3)$ , so $X (s) = 2 - \frac{s + 3}{( s + 1 ) ( s + 2 )}$ .
The constant $2$ is the impulse $2 δ (t)$ , which does not affect $x (0^{+})$ . Apply the IVT to the remainder: $lim_{s \to \infty} \frac{- s ( s + 3 )}{( s + 1 ) ( s + 2 )} = - 1$ .
Check by inversion: $- \frac{s + 3}{( s + 1 ) ( s + 2 )} = - \frac{2}{s + 1} + \frac{1}{s + 2}$ , so $x (t) = 2 δ (t) + (- 2 e^{- t} + e^{- 2 t}) u (t)$ and $x (0^{+}) = - 2 + 1 = - 1$ . Applying $lim_{s \to \infty} s X (s)$ to the undivided $X (s)$ would give $\infty$ .

PRACTICE 52 marks · MSQ

An LTI system has system function $H (s) = \frac{s + 1}{( s - 2 ) ( s + 3 )}$ ; its ROC is not specified. Which of the following statements are true?

(A)
If the system is causal, it is not BIBO stable.
(B)
If the system is BIBO stable, its impulse response is non-zero both for some $t > 0$ and for some $t < 0$ .
(C)
Some choice of ROC makes the system both causal and BIBO stable.
(D)
For the BIBO-stable choice of ROC, $h (t) = \frac{2}{5} e^{- 3 t} u (t) - \frac{3}{5} e^{2 t} u (- t)$ .

Answer and solution

Answer: (A), (B), (D)

Partial fractions: $H (s) = \frac{3/5}{s - 2} + \frac{2/5}{s + 3}$ (cover-up: $\frac{2 + 1}{2 + 3} = \frac{3}{5}$ and $\frac{- 3 + 1}{- 3 - 2} = \frac{2}{5}$ ).
Causal: ROC $Re (s) > 2$ , which excludes the $j ω$ axis, so the system is unstable. A is true. C is false, since causal and stable together would need both poles in the LHP.
Stable: ROC $- 3 < Re (s) < 2$ . The pole at $- 3$ lies left of the ROC and gives $\frac{2}{5} e^{- 3 t} u (t)$ ; the pole at $2$ lies right of it and gives $- \frac{3}{5} e^{2 t} u (- t)$ . D is true.
That $h (t)$ is non-zero on both sides of $t = 0$ , so B is true.

PRACTICE 62 marks · NAT

A system obeys $\frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} + 3 y (t) = 6 u (t)$ with $y (0^{-}) = 0$ and $y^{'} (0^{-}) = 2$ . The value of $y (t)$ at $t = 1$ s, rounded to two decimal places, is ______.

Answer and solution

Answer: $1.26$ (accepted range $\pm 0.01$ )

Unilateral transform: $s^{2} Y - 0 - 2 + 4 (s Y - 0) + 3 Y = \frac{6}{s}$ .
$(s + 1) (s + 3) Y = 2 + \frac{6}{s} = \frac{2 ( s + 3 )}{s}$ , so $Y (s) = \frac{2}{s ( s + 1 )}$ ; the factor $s + 3$ cancels.
$Y (s) = \frac{2}{s} - \frac{2}{s + 1}$ , so $y (t) = 2 - 2 e^{- t}$ for $t \geq 0$ . Check: $y (0) = 0$ and $y^{'} (0) = 2$ .
$y (1) = 2 - 2 e^{- 1} = 2 - 0.7358 = 1.26$ .

PRACTICE 72 marks · MCQ

A causal, stable system has $H (s) = \frac{s - 1}{( s + 2 ) ( s + 3 )}$ . It is written as $H (s) = H_{m i n} (s) H_{ap} (s)$ , where $H_{m i n} (s)$ is minimum phase (causal and stable, all poles and zeros in the open LHP) and $H_{ap} (s)$ is all-pass with $∣ H_{ap} (j ω) ∣ = 1$ for all $ω$ . Which pair is correct?

(A)
$H_{m i n} (s) = \frac{s + 1}{( s + 2 ) ( s + 3 )}$ and $H_{ap} (s) = \frac{s - 1}{s + 1}$
(B)
$H_{m i n} (s) = \frac{s - 1}{( s + 2 ) ( s + 3 )}$ and $H_{ap} (s) = 1$
(C)
$H_{m i n} (s) = \frac{1}{( s + 2 ) ( s + 3 )}$ and $H_{ap} (s) = s - 1$
(D)
$H_{m i n} (s) = \frac{s + 1}{( s - 2 ) ( s - 3 )}$ and $H_{ap} (s) = \frac{( s - 1 ) ( s - 2 ) ( s - 3 )}{( s + 1 ) ( s + 2 ) ( s + 3 )}$

Answer and solution

Answer: (A)

The zero at $s = + 1$ is in the RHP, so $H (s)$ itself is not minimum phase: B is out.
Reflect the zero to $s = - 1$ : $H_{m i n} (s) = \frac{s + 1}{( s + 2 ) ( s + 3 )}$ has every pole and zero in the LHP. The leftover factor is $\frac{s - 1}{s + 1}$ , and $\frac{j ω - 1}{j ω + 1} = \frac{ω ^{2} + 1}{ω ^{2} + 1} = 1$ . The product is $H (s)$ , so A is correct.
C: $∣ j ω - 1∣ = 1 + ω^{2}$ is not constant, so $s - 1$ is not all-pass.
D: the product is $H (s)$ and the second factor is all-pass, but the first factor has poles at $+ 2$ and $+ 3$ , so it cannot be causal and stable; it is not minimum phase.

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