Question 3.SGPYQ.40: Let y[n] denote the convolution of h[n] and g[n], where h[n]......
Let y[n] denote the convolution of h[n] and g[n], where h[n]=(1 / 2)^n \quad u[n] and g[n] is a causal sequence. If y[0] = 1 and y[1] =1/2, then g[1] equals
(a) 0 (b) \frac{1}{2} (c) 1 (d) \frac{3}{2}
Given that
h[n]=\left\lgroup \frac{1}{2} \right\rgroup^n u(n) \quad y[0]=1 \quad y[1]=\frac{1}{2}
Convolution sum,
y[n]=h[n] * g[n]=\sum_{k=-\infty}^{\infty} h(n) g(n-k)
For causal sequence,
\begin{aligned} & y[n]=\sum_{k=0}^{\infty} h(n) g(n-k) \\ & y[n]=h(n) g(n)+h(n) g(n-1)+h(n) g(n-2)+\ldots . \end{aligned}
When n = 0;
\begin{aligned} y[0] & =h(0) g(0)+h(1) g(-1)+\ldots . . \\ & (\because \text { for casual system } g(-1)= g (-2)+\ldots \ldots=0) \\ & =h(0) g(0) \end{aligned}
When n = 1,
\begin{aligned} y[1] & =h(1) g(1)+h(1) g(0)+h(2) g(-1)+\ldots \ldots \\ & =h(1) g(1)+h(1) g(0) \end{aligned}
Since
h[1]=\left\lgroup \frac{1}{2} \right\rgroup^1=\frac{1}{2} \text { and } y[1]=\frac{1}{2}
We have
\frac{1}{2}=\frac{1}{2}[g(1)+g(0)] \Rightarrow g(1)=1-g(0)
Already we have,
y(0)=h(0) g(0) ; g(0)=\frac{g(0)}{h(0)}=\frac{1}{1}=1
Therefore, g(1) = 1-1 0