Question

Consider a sampler which samples the continuous-time input signal x(t) at a sampling frequency fs = 8000 Hz and produces at its output a sampled discrete-time signal x$(t) = x(nTs), where To = 1/fs is the sampling period. If the sampled signal is passed through a unity-gain lowpass filter with cutoff frequency of fs/2, sketch the magnitude spectrum of the resulting signal for the following input signals: (a) x(t) = cos(6000nt). (b) x(t) = cos(12000nt). (c) x(t) = cos(18000nt).

Answer 1

$Sampling\:frequency=f_s=8000Hz=8\,KHz$

$Cutoff\:frequency\:of\:LPF=\frac{f_s}{2}=4000Hz=4KHz$

a)

$x(t)=cos\,6000\pi t=cos(2\pi.3000t)$

$Spectrum \:of\: x(t)=X(F)=\frac{1}{2}(\delta(F-3)+\delta(F+3))\:\:[where\, F \,is \,in\, KHz]$

Spectrum after sampling:

$X_n(F)=\frac{1}{2}(\delta(F-3\pm 8n)+\delta(F+3\pm 8n))\:,n=0,1,2.....$

So, we will get 0.5 magnitude at frequencies:

Putting n=0, F=3,-3

$Putting\: n=1, F=\left\{\begin{matrix} 3-8=-5\\ 3+8=11\\ -3-8=-11\\ -3+8=5 \end{matrix}\right.$

Going on like this, we can draw the spectrum of sampled signal:

After passing through a low pass filter with cutoff frequency d KHz, only the frequency components less than 4 KHz will remain in the spectrum. As the LPF has unity gain, magnitude will remain same. So, the spectrum will be,

(XIF)| 1 -Il-8-5-3 o ६ 5 8 F (KH2)

0.5 0.5 -3 3 F(KHz)

b)

$x(t)=cos\,12000\pi t=cos(2\pi.6000t)$

$Spectrum \:of\: x(t)=X(F)=\frac{1}{2}(\delta(F-6)+\delta(F+6))\:\:[where\, F \,is \,in\, KHz]$

Spectrum after sampling:

$X_n(F)=\frac{1}{2}(\delta(F-6\pm 8n)+\delta(F+6\pm 8n))\:,n=0,1,2.....$

$putting\:n=0,\:F=-6,6$

$Putting\: n=1, F=\left\{\begin{matrix} 6-8=-2\\ 6+8=14\\ -6-8=-14\\ -6+8=2 \end{matrix}\right.$

$Putting\: n=2, F=\left\{\begin{matrix} 6-16=-10\\ 6+16=22\\ -6-16=-22\\ -6+16=10 \end{matrix}\right.$

Spcetrum of sampled signal:

1*(F) 0.5 11 a 2 6 018 ry - 14 70-8-6-2 PF (KH2)

After passing through a low pass filter with cutoff frequency 4 KHz, only the frequency components less than 4 KHz will remain in the spectrum. As the LPF has unity gain, magnitude will remain same. So, the spectrum will be,

0.5 0.5 -2 F(KHz)

c)

$x(t)=cos\,18000\pi t=cos(2\pi.9000t)$

$Spectrum \:of\: x(t)=X(F)=\frac{1}{2}(\delta(F-9)+\delta(F+9))\:\:[where\, F \,is \,in\, KHz]$

Spectrum after sampling:

$X_n(F)=\frac{1}{2}(\delta(F-9\pm 8n)+\delta(F+9\pm 8n))\:,n=0,1,2.....$

$putting\:n=0,\:F=-9,9$

$Putting\: n=1, F=\left\{\begin{matrix} 9-8=1\\ 9+8=17\\ -9-8=-17\\ -9+8=1 \end{matrix}\right.$

$Putting\: n=2, F=\left\{\begin{matrix} 9-16=-7\\ 9+16=25\\ -9-16=-25\\ -9+16=7 \end{matrix}\right.$

Spectrum of sampled signal:

After passing through a low pass filter with cutoff frequency 4 KHz, only the frequency components less than 4 KHz will remain in the spectrum. As the LPF has unity gain, magnitude will remain same. So, the spectrum will be,

(Thaly EI SI 6 t 10- t. 6- Sl-ti- I IGA

0.5 0.5 -1 1 F(KHz)