Sec.2.2 The Physical Layer:Channels and Modems 43 Area≈5s(T) Area≈islr+6) 5t) T+6 8(r)h(t-r) 6sr+6}ht-T-6) Figure 2.5 Graphical interpretation of the convolution equation.Input s(t)is viewed as the superposition of narrow pulses of width 6.Each such pulse yields an output 6s()h(t-7).The overall output is the sum of these pulse responses. time t.Adding the responses at time t from each of these input pulses and going to the limit6→0 shows that r-Jx s(T)h(t -T)dT (2.1) This formula is called the convolution integral,and r(t)is referred to as the con- volution of s(t)and h(t).Note that this formula asserts that the filtering aspects of a channel are completely characterized by the impulse response h(t);given h(t),the out- put r(t)for any input s(t)can be determined.For the example in Fig.2.3,it has been assumed that h(t)is 0 for t<0 and ae-for t0.where =2/T.Given this,it is easy to calculate the responses in the figure. From Eq.(2.1),note that the output at a given time t depends significantly on the input s(7)over the interval where h(t-T)is significantly nonzero.As a result,if this interval is much larger than the signaling interval T between successive input pulses, significant amounts of intersymbol interference will occur. Physically,a channel cannot respond to an input before the input occurs,and therefore h(t)should be 0 for t <0;thus,the upper limit of integration in Eq.(2.1) could be taken as t.It is often useful,however,to employ a different time reference at the receiver than at the transmitter,thus eliminating the effect of propagation delay.In this case,h(t)could be nonzero for t<0,which is the reason for the infinite upper limit of integration. 2.2.2 Frequency Response To gain more insight into the effects of filtering,it is necessary to look at the frequency domain as well as the time domain;that is,one wants to find the effect of filtering.... Sec. 2.2 The Physical Layer: Channels and Modems lis(r)h(t - r) ~IiS(r+Ii)h(t -r-li) r r+1i t-- t-- 43 (2.1 ) Figure 2.5 Graphical interpretation of the convolution equation. Input s(t) is viewed as the superposition of narrow pulses of width 6. Each such pulse yields an output 6S(T)h(t - T). The overall output is the sum of these pulse responses. time t. Adding the responses at time t from each of these input pulses and going to the limit {) ---+ 0 shows that T(t) = l:x s(T)h(t - T)dT This fonnula is called the convolution integral, and T(t) is referred to as the con￾volution of set) and h(t). Note that this fonnula asserts that the filtering aspects of a channel are completely characterized by the impulse response h(t); given h(t), the out￾put ret) for any input set) can be detennined. For the example in Fig. 2.3, it has been assumed that h(t) is 0 for t < 0 and ae- n / for t 2: 0, where a = 21T. Given this, it is easy to calculate the responses in the figure. From Eq. (2.1), note that the output at a given time t depends significantly on the input SeT) over the interval where h(t - T) is significantly nonzero. As a result, if this interval is much larger than the signaling interval T between successive input pulses, significant amounts of intersymbol interference will occur. Physically, a channel cannot respond to an input before the input occurs, and therefore h(t) should be 0 for t < 0; thus, the upper limit of integration in Eq. (2.1) could be taken as t. It is often useful, however, to employ a different time reference at the receiver than at the transmitter, thus eliminating the effect of propagation delay. In this case, h(t) could be nonzero for t < 0, which is the reason for the infinite upper limit of integration. 2.2.2 Frequency Response To gain more insight into the effects of filtering, it is necessary to look at the frequency domain as well as the time domain; that is, one wants to find the effect of filtering