1. Given a digital filter\'s output in the figure 2, and sampling freqcy of 10kH

Question

1. Given a digital filter's output in the figure 2, and sampling freqcy of 10kHz, calculate (approximatcly) the filter's output y(t) for the following input signal 4 2(t) 0.5 * sin (2 * * 2000 * t) + 0.3 * cos(2 * * 3000 * t) 0.8 0.6 04 02 0.5 Normalized Frequency n rad/sample) 0.1 0.2 0.3 0.4 0.8 0.7 0.8 0.9 Figure 2 2. You are given the FIR filter shown in figure 3, and set of coefficients h0- 1. h1 0.h2 0.25. h3-0.5, and h4-0.75 If a sequence of input signals Xn, given in the table that is part of figure 3 is used, calculate the respouse of the filter Yn. [6 Hint: Assume that all registcrs initially contain zcro. Also Xn is fed from left to right i. e. 10 first then 20 and so on. 10 20 10 0 30 40 Figure 3:

Explanation / Answer

let's consider the components of the input.

1. The first component is 0.5*sin(2*pi*2000*t). Its normalised frequency is (2*pi*2000)/(sampling frequency) = (2*pi*2000)/(40*1000) = 0.1*pi.

the magnitude of filter at this normalised frequency is 0.9 (approximately).

the output at this normalised frequency is x(t) * H(w) = 0.5*sin(2*pi*2000*t) * 0.9 = 0.45*sin(2*pi*2000*t)

2. The second component is 0.3*cos(2*pi*3000*t). Its normalised frequency is (2*pi*3000)/(sampling frequency) = (2*pi*3000)/(40*1000) = 0.15*pi.

the magnitude of filter at this normalised frequency is 0.8 (approximately).

the output at this normalised frequency is x(t) * H(w) = 0.3*cos(2*pi*3000*t) * 0.8 = 0.24*cos(2*pi*3000*t).

So, the total output is 0.45*sin(2*pi*2000*t) + 0.24*cos(2*pi*3000*t).

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