By Alan V. Oppenheim, Ronald W. Schafer, John R. Buck

ISBN-10: 0137549202

ISBN-13: 9780137549207

For senior/graduate-level classes in Discrete-Time sign Processing. THE definitive, authoritative textual content on DSP - excellent for people with an introductory-level wisdom of indications and structures. Written via favourite, DSP pioneers, it offers thorough remedy of the basic theorems and houses of discrete-time linear platforms, filtering, sampling, and discrete-time Fourier research. through concentrating on the overall and common suggestions in discrete-time sign processing, it continues to be very important and proper to the recent demanding situations bobbing up within the box -without proscribing itself to precise applied sciences with really brief existence spans.

Uploader Notes:

This is the ideas guide for the second one variation. comprises a number of the difficulties from the 3rd variants.

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**Additional info for Discrete-Time Signal Processing: Solutions Manual (2nd Edition)**

**Sample text**

N], therefore y[n] shifts by no and the system is thus shift invariant. n]. ~ a result, y[n] is bounded and therefore the system is stable. (d) No, the system is not causal. Values of the output at time n depend on values ofthe input at time n +I (through u~[n] and m,[n]). Since present values of the ouput depend of future values ofthe input, the system cannot be causal. [n] is very large, u;[n] is zero, therefore: = y[n] m,[n] l n+l = 3 L: :r[k] i:=n-1 which is the average of the previous, present and next value of the input.

When a zero-mean random process is input to a determistic LTI system, the output is also zeromean: .. L Taking the expected value of both sides: ""' = .... = 0, z[k]h[n- k]. E{z[n]}h[n- k] ifm. = 0. So, u! E L.. h[m]h[k]E{z[n- m]z[n- k]} .. E a! L h'[m]. E .. I'JUII'-oct=-oo = ,.. 86. 85: (a) 00 ~ =~ :E h~[kJ this statement is TRUE, because o:[n] is a white noise sequence. (b) Since y[n] is not a white noise sequence, this statement is FALSE. (c) Let h, [n] = h2[n] = anu[n] bnu[n]. , n = :Ea•bn-•, n;<:O bn ( 1 - (a/W+') u[n] 1 -(a/b) = w[n] = o:[n] • h[n].

3. 4. The pole-zero plot of X(z) appears below. X(z} 3 2 (a) For the Fourier transform of :z:(n] to exist, the z-transform of z[n] must have an ROC which includes the unit circle, therefore, Iii < lzl < 121. Since this ROC lies outside ~, this pole contributes a right-sided sequence. Since the ROC lies inside 2 and 3, these poles contribute left-sided sequences. The overall :z:[n] is therefore two-sided. (b) Two-sided sequences have ROC's which look like washers. There are two possibilities.

### Discrete-Time Signal Processing: Solutions Manual (2nd Edition) by Alan V. Oppenheim, Ronald W. Schafer, John R. Buck

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