Signal Processing - Study Mode
[#276] Match List-I with List-II and select the correct answer using the options given below: List-I List-II a. $${alpha ^n}uleft( n
ight)$$ 1. $$frac{{alpha {z^{ - 1}}}}{{{{left( {1 - alpha {z^{ - 1}}}
ight)}^2}}}{ ext{ROC}}:left| z
ight| > left| alpha
ight|$$ b. $$ - {alpha ^n}uleft( { - n - 1}
ight)$$ 2. $$frac{1}{{left( {1 - alpha {z^{ - 1}}}
ight)}}{ ext{ROC}}:left| z
ight| > left| alpha
ight|$$ c. $$ - n{alpha ^n}uleft( { - n - 1}
ight)$$ 3. $$frac{1}{{left( {1 - alpha {z^{ - 1}}}
ight)}}{ ext{ROC}}:left| z
ight| < left| alpha
ight|$$ d. $$n{alpha ^n}uleft( n
ight)$$ 4. $$frac{{alpha {z^{ - 1}}}}{{{{left( {1 - alpha {z^{ - 1}}}
ight)}^2}}}{ ext{ROC}}:left| z
ight| < left| alpha
ight|$$
Correct Answer
(D) a-2, b-3, c-4, d-1
[#277] The positions of the poles for the Butterworth filter lie on . . . . . . . . and the positions of the poles for the Chebyshev filter lie on . . . . . . . .
Correct Answer
(B) Circle, Ellipse
[#278] A periodic signal x(t) has a trigonometric Fourier series expansion $$xleft( t
ight) = {a_0} + sumlimits_{n = 1}^infty {left( {{a_n},cos ,n{omega _0}t + {b_n}sin ,n{omega _0}t}
ight)} $$ If $$xleft( t
ight) = - xleft( { - t}
ight) = - xleft( {{{t - pi } over {{omega _0}}}}
ight),$$ xa0 xa0xa0 we can conclude that
Correct Answer
(A) a n are zero for all n and b n are zero for n even
[#279] If $$Fleft( s
ight) = Lleft| {fleft( t
ight)}
ight| = {K over {left( {s + 1}
ight)left( {{s^2} + 4}
ight)}},$$ xa0 xa0 xa0 then $$mathop {lim }limits_{t o infty } fleft( t
ight)$$xa0 is given by
Correct Answer
(C) Infinite
[#280] The input to a channel is a bandpass signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by $$yleft( t
ight) = left( {{1 over {100}}}
ight)cos left( {100t - {{10}^{ - 6}}}
ight)cos left( {{{10}^6}t - 1.56}
ight)$$ The group delay (t g ) and the phase delay (t p ) in seconds, of the channel are
Correct Answer
(D) t g = 10 8 , t p = 1.56