InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
The Laplace transform of unit ramp function starting at = is |
| Answer» £f(t) = £-1F(s) = f(t) £[a f1(t) + bf2(t)] = aF1(s) + bF2(s) where £[f(t - T)] = e-sT F(s) £[e-at f(t)] = F(s + a) Initial value theorem Final value theroem Convolution Integral where t is dummy variable for t. | |
| 102. |
Transfer function of a linear system is 200 The system is a |
| Answer» Because H(ω) = 200 e-j10ω And |H(ω)| = 200 with zero phase. So it cannot be attenuator. | |
| 103. |
which one is discrete time periodic signal? |
| Answer» In Discrete time signal irrational function cannot be periodic, frequency must be rational. Hence only (c) is periodic. sin 3pn = sin 3p(n + N) ⇒ for Periodic 3pN = 2Kp . | |
| 104. |
Frequency domain of a periodic triangular function is a |
| Answer» Because F.T. of a Triangular function is square sampling function. | |
| 105. |
The -transform of sequence x [] is |
| Answer» It is differentiation property. | |
| 106. |
The inverse Laplace transform of |
| Answer» . | |
| 107. |
If =0 |
| Answer» . | |
| 108. |
An ac circuit has an impedance of (2 - 9) Ω for third harmonic. The impedance for fundamental is |
| Answer» . | |
| 109. |
The ROC of sequence [] = (0.8) ∪[] + (0.4) ∪[] |
| Answer» ROC ⇒ 1 - 0.8 z-1 < 0 1 - 0.4 z-1 ⇒ 1 > 0.8 z-1 ⇒ z > 0.4 or z > 0.8 By combining both condition ROC because z > 0.4 If you consider z > 0.8 then ROC between the limit 0 to 0.4 excluded. | |
| 110. |
Pick the odd one |
| Answer» Because Variance, standard Deviation, Expectation are related to each other. | |
| 111. |
The sampling of a function () = sin 2 starts from a zero crossing. The signal can be detected if sampling time T is |
| Answer» Because fs ≤ 2f0, Ts ≤ . | |
| 112. |
If then for this to be true () is |
| Answer» If . | |
| 113. |
For the discrete time system of the given figure |
| Answer» uk + 0.5 yk - 1 - 0.25 yk - 2 = yk. | |
| 114. |
The analog signal () is given below () = 4 cos 100 + 8 sin 200 + cos 300 , the Nyquist sampling rate will be |
| Answer» m (t) = 4 cos 100 pt + 8 sin 200 pt + cos 300 pt Nyquist sampling freq fs ≤ 2fm where fm is highest frequency component in given signal and highest fm in 3rd part 2pfmt = 300 pt fm = 150 Hz fs = 2 x 150 p 300 Hz . | |
| 115. |
The ROC of sequence in the Z.T. of sequence [] = ∪ [] is |
| Answer» The simplest method to find the ROC. Put denominator to greater than zero. | |
| 116. |
In Laplace transform, multiplication by in time domain becomes |
| Answer» £e-at f(t) = F(s + a). | |
| 117. |
Laplace transform of cos () is |
| Answer» £f(t) = £-1F(s) = f(t) £[a f1(t) + bf2(t)] = aF1(s) + bF2(s) where £[f(t - T)] = e-sT F(s) £[e-at f(t)] = F(s + a) Initial value theorem Final value theroem Convolution Integral where t is dummy variable for t. | |
| 118. |
The impulse response of discrete time system is [] = (4) ∪[3 - ], the system is |
| Answer» For stability Hence it is stable and it depend upon future value for - ve n so it is non causal. | |
| 119. |
The value of the in + ve sense is |
| Answer» Apply Cauchy integral formula. | |
| 120. |
The amplitude of the first odd harmonic of the square wave shown in the given figure is |
| Answer» Fourier coefficient . | |
| 121. |
The inverse Laplace transform of is |
| Answer» . | |
| 122. |
A voltage wave = 10 + 20 sin ω + 7.5 sin 3 ω() is applied to a series combination of two elements. The current is = 5 sin (ω + 20°) + 1.5 sin (3ω + 10°). The elements are |
| Answer» DC component is absent in current and current is leading the voltage. Hence R and C. | |
| 123. |
If F() = δ( - ), F()= |
| Answer» It is an impulse originating at t = a. | |
| 124. |
The property is not valid for basic singularity function is |
| Answer» Singularity function is discontinuous at origin like ∪(t), δ(t). | |
| 125. |
Magnitude Plot of a Composite signal () = + is |
| Answer» x(t) = e2jt ⇒ x(t) = 2ej2.5t cos(0.5t) The magnitude |x(t)| = 2 cos |0.5t| which is a full rectified wave. | |
| 126. |
The final value of is |
| Answer» . | |
| 127. |
Let () and () with F.T. () and () respectively be related as shown in figure Then () is |
| Answer» By applying time shifting and scaling property. | |
| 128. |
Which one is a linear system? |
| Answer» For linearity y1[n] = x1[n] + x2[n - 10] ...(1) y2 = x2[n] + x2[n - 10] ...(2) y1[n] = x1[n] + x2[n] + x2[n - 10] + x2[n - 10] ...(3) Now find y1[n] + y2[n] Corresponding to x1[n] + x2[n] It is same as equation (3) hence linear. But in part (c) y[n] = x2[n] ⇒ y1[n] = x21[n], y2[n] = x22[n]⇒ y1[n] + y2[n] = x22[n]...3 But y1[n] + y2[n] Corresponing x1[n] + x2[n] is y1[n] + y2[n] = {x1[n] + x2[n]}2 = x12[n] + x22[n] + 2x1[n] x2[n]....4 Equations (3) and (4) are not same hence not linear. | |
| 129. |
Laplace transform of a pulse function of magnitude and duration from = 0 to = is |
| Answer» . | |
| 130. |
The continuous time system with impulse response () = is stable, for is even, when |
| Answer» For stable system exponential must be -ve so that a < 0. | |
| 131. |
If I () , initial value of () is |
| Answer» . | |
| 132. |
F.T. of normalized Gaussian function is |
| Answer» Apply Differential Property of F.T Note that this is true only for e-pt2 If g(t) = e-at2 then G(f) = . | |
| 133. |
If F() is the Laplace transform of () then Laplace transform of |
| Answer» £f(t) = £-1F(s) = f(t) £[a f1(t) + bf2(t)] = aF1(s) + bF2(s) where £[f(t - T)] = e-sT F(s) £[e-at f(t)] = F(s + a) Initial value theorem Final value theroem Convolution Integral where t is dummy variable for t. | |
| 134. |
An ac sinusoidal wave has an rms value of 10 V. The peak to peak value is |
| Answer» Peak to peak value = (rms value). | |
| 135. |
A complex wave is 5 + 5 sin ω. Its rms value is |
| Answer» | |
| 136. |
A function having frequency is to be sampled. The sampling time T should be |
| Answer» Sampling frequency must be more than 2f. Therefore . | |
| 137. |
The final value theorem is |
| Answer» £f(t) = £-1F(s) = f(t) £[a f1(t) + bf2(t)] = aF1(s) + bF2(s) where £[f(t - T)] = e-sT F(s) £[e-at f(t)] = F(s + a) Initial value theorem Final value theroem Convolution Integral where t is dummy variable for t. | |
| 138. |
If |
| Answer» det.|A| = 3(8 - 12) = - 12. | |
| 139. |
Inverse Fourier transform of sgn (ω) is |
| Answer» By duality Property. | |
| 140. |
If I () , the final value of () is |
| Answer» | |
| 141. |
A signal () is multiplied by a sinusoidal waveform of frequency such that()=() cos 2If Fourier transform of () is M(), Fourier transform of () will be |
| Answer» It is a modulation process. The resultant has (f + fc) and (f - fc) terms. | |
| 142. |
A voltage wave having 5% fifth harmonic content is applied to a series RC circuit. The percentage fifth harmonic content in the current wave will be |
| Answer» IC = V(jωC). | |
| 143. |
δ() is a |
| Answer» Because Unit step is a Power signal. So By trignometric identifies d(t) also power. . | |
| 144. |
The analog signal given below is sampled by 600 samples per second for () = 3 sin 500 + 2 sin 700 then folding frequency is |
| Answer» . | |
| 145. |
The signal defined by the equations () = 0 for < 0, () = E for 0 ≤ ≤ and () = 0 for > is |
| Answer» It is a pulse lasting for t = a. | |
| 146. |
Inverse Laplace transform of is |
| Answer» f(t) = e-2t + e-3t. | |
| 147. |
Let be the Fourier transform of (), then () is |
| Answer» . | |
| 148. |
The Laplace transform of a vlotage across a capacitor is V () = . If capacitor is 0.5 F, The current through capacitor at = 0 is |
| Answer» . | |
| 149. |
The value of Integral ( + 2) δ( - 3) is equal to |
| Answer» Because t = 3 is lies out of integral limit. | |
| 150. |
The transform of sequence [] = {2, 4, 3, 2} |
| Answer» n(z) = 2z+2 + 4z+1 + 3 + 2z-1 or 2z-1 + 3 + 4z+1 + 2z+2 . | |