Laplace formulas

Laplace transform, poles, zeros and system functions. 25 formulas for GATE EC, EE, IN Signals and Systems, each with when to use it. Each is explained in the topic notes.

All 25 Laplace formulas

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.

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 $s$ ).

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 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 $e^{- s t_{0}}$ .

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.

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.

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 $2∣ K ∣ e^{- a t} cos (b t + ∠ K) u (t)$ .

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 $K$ .

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 $F$ .

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 $K < 0$ .

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.

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 $X (s)$ .

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.

Where this was asked: 2020 (2), 2022 (1), 2023 (1), 2024 (2), 2026 (2). See the 8 previous-year questions on Laplace. This shows what was asked, not what will be asked.

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Laplace: the notes and 8 solved previous-year questions