600 
R(t) 
es 
le} 0.2 0.4 0.6 08 1.0 1.2 1.4 16 18 2.0 2.2 2.4 26 28 3.0 
1/T 
Figure 18 - Response of an Amplifier to an Input Signal of the Form elt 
The amplifier in question is one which shows the response to a unit pulse depicted by the 
lower curve in Figure 16. The response is expressed by Equation [17] 
and is plotted for two values of t,/T. 
/t ‘ 1 /T ( y/ 
R(t) = 1 — e Mo — fe IM(y — ROM lar {16] 
T 
0 
a 1h -t/T -t/t, 
R(t) = Tt ( € ) [17] 
See Reference (5). 
The original signal S(t) and response R(t) are plotted for the in- 
dicated values of t,/T in Figure 18. It is apparent that the response ap- 
proaches the input signal more closely as t,/T approaches zero. There is a 
considerable loss of peak signal even for small values of toy for example, 
a loss of 7.5 per cent for t,/T= 1/50. 
It is useful to obtain the peak response as a function of t/t. The 
t,/T necessary for a given allowable loss of peak response for the assumed 
type of signal can then be specified. Differentiating Equation [17] yields 
, Ti [ type 1 i 
=>lclUlU CT — — + — 
R'(t) T to € ty € 
Setting this expression equal to zero, we obtain 
