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sec5solns - ECE 2200 Signals and Information Section 5(Week...

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ECE 2200 Signals and Information – Section 5 (Week 5 Mar 1-4) Page 1 1. The FIR filter with impulse response h [ n ] = 2 δ [ n - 2] has output y [ n ] = u [ n - 3] - u [ n - 6] . Determine the input x[n]. ————————————— ± ————————————— Solution (see “Section 5-6.2.1 Convolution as an Operator”): The deconvolution of y [ n ] is as follows: ± ± ± y [ n ] = x [ n ] * h [ n ] ± ± ± = 2 x [ n - 2] ± ± ± x [ n ] = 1 2 u [ n - 1] - 1 2 u [ n - 4] 2. The FIR filter has a step starting at sample zero as its input, i.e. x [ n ] = u [ n ], and the output is the Kronecker delta function, i.e. y [ n ] = δ [ n ]. Determine the impulse response of the filter. ————————————— ± ————————————— Solution : The output y [ n ] = δ [ n ] is achieved by convolving the unit-step input x [ n ] = u [ n ] with the first difference h [ n ] = δ [ n ] - δ [ n - 1] . 3. Consider the linear, time-invariant system described by the difference equation y [ n ] = 21 x [ n ] - 37 x [ n - 1] + 21 x [ n - 2] (i) Find the frequency response of H ( e j ˆ ω ) as a mathematical formula in polar form. (ii) Determine the period of H ( e j ˆ ω ) which is a periodic function of ˆ ω . (iii) Plot the magnitude and phase of H ( e j ˆ ω ) as a function of ˆ ω for - π < ˆ ω < 5 π . Check your answer using the MATLAB function freqz . (iv) Find ALL frequencies ˆ ω , if any, for which the output response to the input e j ˆ ωn is zero. (v) For the input x [ n ] = sin(2 πn/ 17) determine the output in the form y [ n ] = M cos( γn + θ ) by specifying M , γ , and θ .
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Page 2 ————————————— ± ————————————— Solution (read “Section 6-1 Sinusoidal Response of FIR Systems”): (i) The filter coefficients are b 0 = 21, b 1 = - 37 and b 2 = 21. By Equation (6.4), the frequency-response function (or frequency-response) for the system after factoring out e - j ˆ ω and reducing complex exponentials using Euler’s formula is ± ± ± H ( e j ˆ ω ) = e - j ˆ ω (42 cos ˆ ω - 37) (ii) 2 π (iii) As indicted in the MATLAB documentation, freqz returns a frequency response h and frequency vector w with values ranging from 0 to π . By interpreting this result, we obtain the magnitude and phase plot shown in Figure 1. 0 20 40 60 80 3 2 1 0 1 2 3 - π 0 π 2 π 3 π 4 π 5 π - π 0 π 2 π 3 π 4 π 5 π | H ( e j ˆ ω ) | H ( e j ˆ ω ) Figure 1: (top) Magnitude and (bottom) phase for H ( e j ˆ ω ) over three periods. (iv)
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This note was uploaded on 03/11/2010 for the course ECE 2200 taught by Professor Johnson during the Spring '05 term at Cornell.

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sec5solns - ECE 2200 Signals and Information Section 5(Week...

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