Question 1 |
The period of the discrete-time signal x[n] described by the equation below is N= ___ (Round off to the nearest integer).
x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)
x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)
16 | |
32 | |
48 | |
64 |
Question 1 Explanation:
Given :
x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)
\begin{aligned} \omega_{1}&=\frac{15 \pi}{8} \\ \therefore \quad \mathrm{N}_{1}&=\frac{2 \pi}{\omega_{1}} \mathrm{~K}-=\frac{2 \pi}{15 \pi} \times 8 \times \mathrm{K}=16 \\ \text { and } \quad \mathrm{N}_{2}&=\frac{2 \pi}{\omega_{2}} \mathrm{~K}=\frac{2 \pi}{\pi} \times 3 \times \mathrm{K}=6 \\ \therefore \quad \text { Period }&=\operatorname{LCM}\left(\mathrm{N}_{1}, \mathrm{~N}_{2}\right) \\ &=\operatorname{LCM}(16,6) \\ &=48 \end{aligned}
x[n]=1+3 \sin \left(\frac{15 \pi}{8} n+\frac{3 \pi}{4}\right)-5 \sin \left(\frac{\pi}{3} n-\frac{\pi}{4}\right)
\begin{aligned} \omega_{1}&=\frac{15 \pi}{8} \\ \therefore \quad \mathrm{N}_{1}&=\frac{2 \pi}{\omega_{1}} \mathrm{~K}-=\frac{2 \pi}{15 \pi} \times 8 \times \mathrm{K}=16 \\ \text { and } \quad \mathrm{N}_{2}&=\frac{2 \pi}{\omega_{2}} \mathrm{~K}=\frac{2 \pi}{\pi} \times 3 \times \mathrm{K}=6 \\ \therefore \quad \text { Period }&=\operatorname{LCM}\left(\mathrm{N}_{1}, \mathrm{~N}_{2}\right) \\ &=\operatorname{LCM}(16,6) \\ &=48 \end{aligned}
Question 2 |
For the signals x(t) and y(t) shown in the figure, z(t)=x(t) * y(t) is maximum at t=T_{1}. Then T_{1} in seconds is (Round off to the nearest integer).
2 | |
3 | |
4 | |
5 |
Question 2 Explanation:
Using convotution property
\begin{aligned} z(t) & =x(t) * y(t) \\ & =\int_{-\infty}^{\infty} \mathrm{y}(\tau) \mathrm{x}(\mathrm{t}-\tau) \mathrm{d} \tau \end{aligned}
For max. overlap
\begin{aligned} \therefore \quad -1+t&=3\\ \Rightarrow \quad t&=4 sec \\ or, \quad \quad T_1&=4 sec \end{aligned}
\begin{aligned} z(t) & =x(t) * y(t) \\ & =\int_{-\infty}^{\infty} \mathrm{y}(\tau) \mathrm{x}(\mathrm{t}-\tau) \mathrm{d} \tau \end{aligned}
For max. overlap
\begin{aligned} \therefore \quad -1+t&=3\\ \Rightarrow \quad t&=4 sec \\ or, \quad \quad T_1&=4 sec \end{aligned}
Question 3 |
Two discrete-time linear time-invariant systems with impulse responses h_{1}\left [ n \right ]=\delta \left [ n-1 \right ]+\delta \left [ n+1 \right ] and h_{2}\left [ n \right ]=\delta \left [ n\right ]+\delta \left [ n-1 \right ] are connected in cascade, where \delta \left [ n\right ] is the Kronecker delta. The impulse response of the cascaded system is
\delta \left [ n-2\right ]+\delta \left [ n+1 \right ] | |
\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ] | |
\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ] | |
\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ] |
Question 3 Explanation:
\begin{aligned} h(n) &=\text { Resultant impulse response } \\ &=h_{1}(n) * h_{2}(n) \end{aligned}
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)
Question 4 |
Suppose for input x(t) a linear time-invariant system with impulse response h(t) produces
output y(t), so that x(t)*h(t)=y(t). Further, if |x(t)|*|h(t)|=z(t), which of the following
statements is true?
For all t \in (-\infty ,\infty ), z(t)\leq y(t) | |
For some but not all t \in (-\infty ,\infty ), z(t)\leq y(t) | |
For all t \in (-\infty ,\infty ), z(t)\geq y(t) | |
For some but not all t \in (-\infty ,\infty ), z(t)\geq y(t) |
Question 4 Explanation:
\begin{aligned}
\text{Since, } y(t) &= x(t) + h(t)\\ \text{and }z(t)&=\left | x(t) \right |\times \left | h(t) \right |
\end{aligned}
Case-1:
case 2:
\text{then, }y(t)=z(t)
\text{Thus, } z(t)\geq y(t), \text{ for all 't'}
Case-1:
case 2:
\text{then, }y(t)=z(t)
\text{Thus, } z(t)\geq y(t), \text{ for all 't'}
Question 5 |
The symbols, a and T, represent positive quantities, and u(t) is the unit step function. Which one of the following impulse responses is NOT the output of a causal linear time-invariant system?
e^{+at}u(t) | |
e^{-a(t+T)}u(t) | |
1+e^{-at}u(t) | |
e^{-a(t-T)}u(t) |
Question 5 Explanation:
a and T represents positive quantities.
u(t) is unit step function.
h(t)=1+e^{-at}u(t), is non-causal
Here '1' is a constant and two sided so the system will be non-causal, because for causal system,
h(t)=0,\;\;\; t \lt 0
h(t) \neq 0,\;\;\; t \gt 0
u(t) is unit step function.
h(t)=1+e^{-at}u(t), is non-causal
Here '1' is a constant and two sided so the system will be non-causal, because for causal system,
h(t)=0,\;\;\; t \lt 0
h(t) \neq 0,\;\;\; t \gt 0
There are 5 questions to complete.