Question 1 |
The Z-transform of a discrete signal x[n] is
X(z)=\frac{4z}{\left ( z-\frac{1}{5} \right ) \left ( z-\frac{2}{3} \right ) (z-3)} with ROC = R
Which one of the following statements is true?
X(z)=\frac{4z}{\left ( z-\frac{1}{5} \right ) \left ( z-\frac{2}{3} \right ) (z-3)} with ROC = R
Which one of the following statements is true?
Discrete-time Fourier transform of x[n] converges if R is |z| \gt 3 | |
Discrete-time Fourier transform of x[n] converges if R is \frac{2}{3} \lt |z| \lt 3 | |
Discrete-time Fourier transform of x[n] converges if R is such that x[n] is a leftsided sequence | |
Discrete-time Fourier transform of x[n] converges if R is such that x[n] is a rightsided sequence |
Question 1 Explanation:
Z = 1, unit circle include by ROC : \frac{2}{3} \lt |z| \lt 3.
Therefore DIFT of x[n] conveys for this ROC.
Therefore DIFT of x[n] conveys for this ROC.
Question 2 |
The casual signal with z-transformer Z^{2}\left ( Z-a \right )^{-2} is
(u[n] is the unit step signal)
(u[n] is the unit step signal)
a^{2n}u\left [ n \right ] | |
\left ( n+1 \right )a^{n}u\left [ n \right ] | |
n^{-1}a^{n}u\left [ n \right ] | |
n^{2}a^{n}u\left [ n \right ] |
Question 2 Explanation:
As we know,
\begin{aligned} n \cdot a^{n} u(n) & \rightleftharpoons \frac{a z}{(z-a)^{2}} \\ \frac{1}{a} \cdot n \cdot a^{n} u(n) & \rightleftharpoons \frac{z}{(z-a)^{2}} \\ f(n) &=n \cdot a^{n-1} u(n) \rightleftharpoons \frac{z}{(z-a)^{2}}=F(z) \end{aligned}
Time-shifting property,
\begin{aligned} f(n) &\rightleftharpoons F(z) \\ f(n+1) &\rightleftharpoons z \cdot F(z) & \\ x(n)&=(n+1) a^{n} u(n+1) \rightleftharpoons \frac{z^{2}}{(z-a)^{2}}=X(z) & \\ \text{Thus},\qquad \qquad x(n)&=(n+1) a^{n} u(n+1)=(n+1) a^{n} \cdot u(n) \\ &[\because(n+1) u(n+1)=(n+1) u(n) \end{aligned}
\begin{aligned} n \cdot a^{n} u(n) & \rightleftharpoons \frac{a z}{(z-a)^{2}} \\ \frac{1}{a} \cdot n \cdot a^{n} u(n) & \rightleftharpoons \frac{z}{(z-a)^{2}} \\ f(n) &=n \cdot a^{n-1} u(n) \rightleftharpoons \frac{z}{(z-a)^{2}}=F(z) \end{aligned}
Time-shifting property,
\begin{aligned} f(n) &\rightleftharpoons F(z) \\ f(n+1) &\rightleftharpoons z \cdot F(z) & \\ x(n)&=(n+1) a^{n} u(n+1) \rightleftharpoons \frac{z^{2}}{(z-a)^{2}}=X(z) & \\ \text{Thus},\qquad \qquad x(n)&=(n+1) a^{n} u(n+1)=(n+1) a^{n} \cdot u(n) \\ &[\because(n+1) u(n+1)=(n+1) u(n) \end{aligned}
Question 3 |
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
stable, real, allpass | |
unstable, complex, allpass | |
unstable, real, highpass | |
stable, complex, lowpass |
Question 3 Explanation:
Since pole zero plot of given transfer function
Since, given pole zero is symmetrical about origin hence it is a all pass system.
Since, one pole on the RHS thus unstable.
Since, pole doesnot have complex conjugate poles and zeros present thus system is not real means system is complex.
Since, given pole zero is symmetrical about origin hence it is a all pass system.
Since, one pole on the RHS thus unstable.
Since, pole doesnot have complex conjugate poles and zeros present thus system is not real means system is complex.
Question 4 |
Consider a signal x[n]=\left ( \frac{1}{2} \right )^n \; 1[n], where 1[n]=0 if n \lt 0, and 1[n]=1 if n \geq 0. The
z-transform of x[n-k], k \gt 0 is \frac{z^{-k}}{1-\frac{1}{2}z^{-1}}
with region of convergence being
|z| \lt 2 | |
|z| \gt 2 | |
|z| \lt 1/2 | |
|z| \gt 1/2 |
Question 4 Explanation:
\begin{aligned}x(n)&=\left (\frac{1}{2} \right )^{n} u(n)
, \; \; \; \text{ROC of }x(n):\left | z \right | \gt \frac{1}{2} \\ x(n-k)\rightleftharpoons X(z)&=\frac{z^{-k}}{1-\frac{1}{2}z^{-1}}
, \; \; \; \text{ROC of }x(n-k): \left | z \right | \gt \frac{1}{2}\\ \text{For } x(n-k) \; \; \; &\text{ROC will be } \left | z \right |\gt \frac{1}{2}\end{aligned}.
Question 5 |
A cascade system having the impulse responses
h_1(n)=\{\overset{1}{\uparrow},-1\} and
h_2(n)=\{\overset{1}{\uparrow},1\}
is shown in the figure below, where symbol \uparrow denotes the time origin.
The input sequence x(n) for which the cascade system produces an output sequence y(n)=\{\overset{1}{\uparrow},2,1,-1,-2,-1\} is
h_1(n)=\{\overset{1}{\uparrow},-1\} and
h_2(n)=\{\overset{1}{\uparrow},1\}
is shown in the figure below, where symbol \uparrow denotes the time origin.
The input sequence x(n) for which the cascade system produces an output sequence y(n)=\{\overset{1}{\uparrow},2,1,-1,-2,-1\} is
x(n)=\{\overset{1}{\uparrow},2,1,1\} | |
x(n)=\{\overset{1}{\uparrow},1,2,2\} | |
x(n)=\{\overset{1}{\uparrow},1,1,1\} | |
x(n)=\{\overset{1}{\uparrow},2,2,1\} |
Question 5 Explanation:
Now,
\begin{aligned} h(n)&=\text{overall system impulse response}\\ &=h_1(n)*h_2(n)\\ &=\{1,-1\}*\{1,1\}\\ &=\{ 1,0,-1\}\underset{*}{\rightleftharpoons }H(z)=1-z^2\\ &\text{As we know,}\\ H(z)&=\frac{Y(z)}{X(z)}\Rightarrow X(z)=\frac{Y(z)}{H(z)}\\ &=\frac{1+2z^{-1}+z^{-2}-z^{-3}-2z^{-4}-z^{-5}}{1-z^{-2}}\\ &=1+2z^{-1}+2z^{-2}+z^{-3}\\ \Rightarrow \; x(n)&=\{1,2,2,1\} \end{aligned}
There are 5 questions to complete.