信号与系统论文
By considering the effect of the superposition sum on each
individual output sample ,we obtain another very useful way to visualize the calculation of y[n]using the convolution sum .In
particular ,consider the evaluation of the output value at some specific time n。A particularly convenient way of displaying this calculation graphically begins with the two signals x[k] and h[n-k]viewed as functions of k .Multiplying these two functions ,we obtain a sequence g[k]=x[k]h[n-k] ,which ,at each time k ,is seen to represent the contribution of x[k] to the output at time n .We conclude that summing all the samples in the sequence of g[k] yields the output value at the selected time n .Thus , to calculate y[n] for all values of n requires repeating this procedure for each value of n .Fortunately ,changing the value of n has a very simple graphical interpretation for the two signals x[k] and h[n-k] ,viewed as functions of k .The following examples illustrate this and the use of the aforementioned viewpoint in evaluating convolution sums .
EXAMPLE 2 .2
Let us consider again the convolution problem encountered in Example 2 .1 .The sequence x[k] is shown in Figure 2 .4(a) ,while the sequence h[n-k] ,for n fixed and
viewed as a function of k ,is shown in Figure 2 .4(b)for several different values of
n .In sketching these sequences ,we have used the fact that h[n-k]is a time-reversed and shifter version of the impulse response h[k] . In particular ,as k increases ,the argument n-k decreases ,explaining the need to perform a time reversal of h[k] .Knowing this ,then in order to sketch the signal h[n-k] ,we need only determine its value k=n . Thus ,if we sketch the signal h[n-k] ,we can obtain the signal h[n-k] simply by shifting to the right if n is positive or to the left if n is negative .the result for our example for values of n<0 ,
n=0 ,1 ,2 ,3 ,and n>3 are shown in Figure 2 .4b Having sketch x[k] and h[n-k] for any particular value of n , we multiply these two signals and sum over all values of k . For our example ,for n<0 ,we see from
Figure2 .4 that x[k]h[n-k]=0 for all k , since the nonzero values of x[k] and h[n-k] do not overlap .consequently , y[n]=0 for n<<0 . For n =0 , since the product of the sequence x[k] with the sequence h[0-k] has only one nonzero sample with the value 0 .5 , we conclude that
(2 .9)
the product of the sequence x[k] with the sequence h[1-k] has two nonzero samples , which may be summed to obtain
(2 .10) similarly
(2 .11) And 2X[k] 0.5 k
2X[k]1H[n-k] n<0 knn-2n-10 1H0-k] -20-1k 1H[1-k] -110k 1H[2-k] 012k 1H[3-k] 012k3 1H[n-k] nn-2n-10k
(2 .12)
Finally ,for n>3 ,the product g[k]=x[k]h[n-k]is zero for all k , from which we conclude that y[n]=0 for n>3 . .The resulting output values agree with those obtained in Example 2 .1 .
信号与系统论文
