GATE 2020 EC Q.5: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EC (IIT Delhi)

1. By definition, the impulse response is the output when the input is $x[n]=\delta [n]$. Here $|x[k]|=|\delta [k]|$ equals $1$ at $k=0$ and $0$ at every other $k$.

2. For $n<0$: the window $-\infty \le k\le n$ never reaches $k=0$, so every $|\delta [k]|$ in it is $0$. Hence $y[n]=\max \{0,0,\dots \}=0$.

3. For $n\ge 0$: the window now contains $k=0$, where $|\delta [0]|=1$, and every other sample in it is $0$. Hence $y[n]=\max \{\dots ,0,1,0,\dots \}=1$.

4. Combining the two cases: $$h[n]=\{\begin{array}{ll}0,& n<0\\ 1,& n\ge 0\end{array}=u[n].$$

5. Checking the other options: (A) is wrong because from $n=0$ onwards the window contains the sample of value $1$. (B) is wrong because for $n<0$ that sample is not yet in the window. (D) is wrong because a running maximum can never decrease: once it has seen the $1$ at $k=0$, it stays at $1$ for every $n\ge 0$ and does not fall back to $0$ at $n=1$.

6. Note: the system is non-linear (because of $|\cdot |$ and $\max$), so $h[n]$ does not fully describe it and cannot be used in a convolution. The question only asks for the output when the input is $\delta [n]$, and that output is $u[n]$.