1.
x(t) = 1, |t|<0.5
x(t) = rect(t) <–––> X(?) = Sa(?/2) or, X(f) = Sa(?f) = sinc(f), sinc(x) = sin(x)/x
H(f) = rect(f/2) or, H(?) = rect(?/4?)
Y(?) = X(?).H(?) = Sa(?/2), |?| < 2?
Y(f) = Sa(?f), |f| < 1
or, Y(f) = sinc(f), |f| < 1,
2.
x(t) = 1, |t|<0.5
x(t) = rect(t) <–––> X(?) = Sa(?/2) or, X(f) = Sa(?f) = sinc(f), sinc(x) = sin(x)/x
H_LPF(f) = rect(f/2) or, H(?) = rect(?/4?)
Y_LPF(?) = X(?).H(?) = Sa(?/2), |?| < 2?
Y_LPF(f) = Sa(?f), |f| < 1
or, Y_LPF(f) = sinc(f), |f| < 1,
H_HPF(f) = 1 - rect(f/2) or, H(?) = 1 - rect(?/4?)
Y_HPF(?) = X(?).H(?) = Sa(?/2), |?| > 2?
Y_HPF(f) = Sa(?f), |f| > 1
or, Y_HPF(f) = sinc(f), |f| > 1,
H_BSF(f) = 1 - rect(|f-5|/2) or, H(?) = 1 - rect(|?-10?|/4?)
Y_HPF(?) = X(?).H(?) = Sa(?/2), |?| > 2?
Y_HPF(f) = Sa(?f), |f| > 1
or, Y_HPF(f) = sinc(f), |f| > 1,
1- The signal x(t) is applied to a low pass filter with cutoff frequency equal to...
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The following periodic signal is input to an ideal low pass filter of bandwidth 25 KHz. 1. x(t) 2 a) Determine the average power of the signal x(t). b) If T 0.1 ms, give the output of the filter as a function of time, y(t) e) Determine the average power of the signal y(t) d) Determine the bandwidth of the signal y(), considered as a baseband signal. e) Now assume that the signal x() (with T-0.1 ms) is instead input...
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