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# Solved objective questions (GATE)

by on May 17, 2009

Using the services of a new author ‘RV’, we are starting a new series of articles in the blog. Typically in India, many of the competitive examinations pertaining to Engineering (GATE, IES) and rectuitment by private and public sector companies (ISRO, BSNL, BEL, BHEL) uses examination with objective questions for the first level screening. We are hoping that this Objective Question Answer series, mostly discussing topics pertaining to Electronics and Communication Engineering, will help those preparing for those examination. Kindly do give your feedback via comment and/or email. Thanks, Krishna

### Q#1: SQNR with pulse code modulation

If the number of bits per sample in a Pulse Coded Modulation (PCM) system is increased from 5 bits to 6 bits,the improvement in signal to quantization noise ratio will be (a) 3 dB (b) 6 dB (c) 2pi dB (d) 0 dB

Explanation

In a pulse coded modulation system, the analog voltage is quantized into discrete voltage. In the post on Signal to Quantization Noise Ratio (SNR) for a quantized sinusoidal, we have derived that the

$SQNR_{dB} = 6.02b + 1.76$, where $b$ is the number of bits in the pulse coded modulation system.

So if the number of bits in a pulse coded modulation system (aka ADC) is increased by 1, the improvement in SQNR is around 6dB. Right answer (b).

### Q#2 Image frequency in superheterodyne receiver

A superheterodyne radio receiver with an intermediate frequency of 455 kHz is tuned to a station operating at 1200 KHz. The associated image frequency is ________ kHz

(a) 1200 (b) 2110 (c) 1100 (d) 455

Explanation

The simple structure of an RF mixer used in a super hetero-dyne receiver is shown below.

Figure: RF mixer in a superheterodyne receiver

Lets assume the following:

$f_{RF}$ is the RF frequency,

$f_{LO}$ is the local oscillator frequency and

$f_{IF}$is the intermediate frequency.

We know, from basic trigonometric identities that the output at the mixer will consist of spectrum at $\pm\left(f_{RF}+f_{LO}\right)$ and $\pm\left(f_{RF}-f_{LO}\right)$.

Note:

We can safely assume that the the $\pm\left(f_{RF}+f_{LO}\right)$ is filtered away. The output of the mixer is a real signal and hence have both +ve and -ve spectrum. Please refer to the post on Negative frequency for bit more details on spectrum of real and complex sinusoidals.

In our example, we have $f_{RF}=1200kHz$ and with $f_{IF}=455kHz$. The possible choices of LO frequency are

(a) $f_{LO}=745kHz$, if $f_{LO}

(b) $f_{LO}=1655kHz$, if $f_{LO}>f_{RF}$

Now, if we assume that

(a) $f_{LO}=745kHz$, its easy to guess that both $f_{RF}=1200kHz$ and $f_{RF}=290kHz (745-455)$ can result in $f_{IF}=455kHz$

(b) $f_{LO}=1655kHz$, its easy to guess that both $f_{RF}=1200kHz$ and $f_{RF}=2100kHz (1655-455)$can result in $f_{IF}=455kHz$.

So the image frequencies are $f_{IM}=290kHz$ $f_{LO} and $f_{IM}=2100kHz$ if $f_{LO}>f_{RF}$.

Given that we do not have 290kHz as a choice, its reasonable to assume that the questionnaire assumed $f_{LO}>f_{RF}$ and hence $f_{IM}=2100kHz$. Right answer : (b)

### Q#3 Bandwidth of wide band FM

A 10 MHz carrier is frequency modulated with a sinusoidal signal at 500Hz, the maximum frequency deviation being 50 KHz. The bandwidth required as given by the Carson’s rule is

(a) 101kHz (b) 50kHz (c) 105 kHz (d) 100kHz

Explanation

In an FM transmission if we send a a sinusoidal $\cos(2pi f_m t)$ at frequency $f_m$is

$s(t) = \cos \left[ 2\pi f_ct+\frac{\Delta f}{f_m}\sin(2pi f_mt) \right]$, where $\Delta f$ is the maximum frequency deviation caused by the message signal.

With some sophisticated mathematical tricks, the signal can be represented as, $s(t) = J_0(\beta) \cos (2pi f_c t) + \\ \sum_{k=1}^{\infty}J_{2k}(\beta)\left[ \sin(2pi (f_c + 2kf_m)t) + \sin(2pi (f_c - 2kf_m)t)\right] + \\\sum_{k=0}^{\infty}J_{2k+1}(\beta)\left[ \cos(2pi (f_c + (2k+1)f_m)t) - \cos(2pi (f_c - (2k+1)f_m)t)\right]$,

where $\beta = \frac{\Delta f}{f_m}$ is called the modulation index, and

$J_n(\beta)$ is the Bessel function of first kind for the value $\beta$.

Strictly, the spectrum of the FM modulated signal has components at frequencies $f_c \pm kf_m$ where $k$ is an integer from 0 to $+\infty$.

However, the value of $J_n(\beta)$ tends to 0 as $k$.

Narrow band FM

If $\beta<<1$, then $J_0(\beta)\approx1$, $J_1(\beta) \approx frac{\beta}{2}$, and $J_k(\beta) \approx 0, k > 1$

Then the FM signal can be approximated as,

$s(t) = J_0(\beta) \cos (2pi f_c t) + J_{1}(\beta)\left[ \cos(2pi (f_c + f_m)t) - \cos(2pi (f_c - f_m)t)\right]$.

This type of FM signal, also called as narrow band FM requires bandwidth of $2f_m$.

Wideband FM

In a wideband FM, the modulation index $\beta \gg 1$ i.e $\Delta f \gg f_m$. Its intuitiveve to guess that we require at-least $2\Delta f$ as the bandwidth. Further, according to Carson’s rule bandwidth of FM in the scenario is given by $2(\Delta f + f_m)$.

So, for our question, the bandwidth is 2*(50+0.5) = 101kHz. Right answer (a)

### Q#4 Aliases with sampling

A 1.0 kHz signal is sampled at the rate of 1.8 kHz and the samples are applied to an ideal rectangular LPF with cut-off frequency of 1.1 kHz, then the output of the filter contains

(a) Only 800 Hz component (b) 800 Hz and 900 Hz components

(c) 800 Hz and 100 Hz components (d) 800Hz, 900 and 100 Hz components

Explanation

According to Nyquist sampling theorem, a bandlimited signal should be sampled atleast greater than the twice of the maximum signal. If this condition is not satisfied aliasing will take place. Figure: Aliasing caused due to under sampling

In the above question, 1.0 kHz signal should be sampled at least at the rate of 2kHz. In this example the sampling was performed at 1.8kHz, whih means the frequency range which does not cause aliasing is from -0.9kHz to +0.9kHz. Any frequency outside this range will get aliased to this range.

In general, if the frequency $f_m$ and the sampling frequency is $f_s$, the aliased signal will be at $f_m \pm kf_s$ .

In this example, the frequency at 1kHz will get aliased to 1-1.8 = -0.8kHz and the frequency at -1kHz will get aliased to -1+1.8 = 0.8kHz. After low pass filtering by 1.1kHz, only 0.8kHz component remains. Right answer (a)

### Q#5: Sampling Double sideband suppressed carrier signal

A 4 GHz carrier is DSB Sc modulated by a low pass message signal with maximum frequency of 2 MHz. The resultant signal is to be ideally sampled. The minimum frequency of the sampling in train should be

(a) 4 MHz (b) 8 MHz (c) 8 GHz (d) 8.004 GHz

Explanation

According to Nyquist sampling theorem, a bandlimited signal should be sampled equal to greater than the twice of the maximum frequency component of the signal. In this example, the spectrum after up-conversion has a maximum frequency of 4000MHz + 2MHz = 4002MHz. So the sampling frequency required to prevent aliasing 8004MHz = 8.004GHz. Right answer (d)

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RNM October 25, 2012 at 2:21 pm

The answer for Q5 is not correct. The signal bandwidth being 2 MHz the AM-DSB has 4 MHz bandwidth. According to bandpass sampling theorem, it is sufficient to sample this signal at 8 MHz

Krishna Sankar October 27, 2012 at 6:51 pm

@RNM: Hmm… i can understand your perspective. Let me re-think.

Aditya August 6, 2012 at 1:45 pm

Hey i am confused with the last question you have given, wouldn’t we be considering it as the bandpass signal and sampling with f(s)=2[f(m)-f(l)]

Krishna Sankar August 9, 2012 at 6:16 am

@Aditya: Hmm… but, the question says “ideally sampled”

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