202 MEASUREMENT OF PRESSURES 



If the input signal is again a negative exponential P{t) = Pme~^/^, 

 the transient response R(t) is given by 



(5.15) 



m) = f^^ [1 - (9/x)e<'-x>»] 



This expression is plotted in Fig. 5.25 as a function of t/d for the case 

 6 = X/10 (circuit time constant a factor of ten greater than the pulse 

 time constant), and also for d/\ = 0, corresponding to perfect low- 



1.0 



0.8 



0.2 



0.8 



1-2 



a. 0.6 

 <n 



Ui 



a: 



0.4 



1.6 



'^'o 12 3 4 3 6 7 8 



REDUCED TIME ^/t^. 



Fig. 5.25 Effect of amplifier low-frequency cutoff on an exponential. 



frequency response. It is seen that the response curve has the same 

 general form as the true curve but lies increasingly below it at increas- 

 ingly longer times. The values of indicated response thus became in- 

 creasingly in error, and are determined as much or more by the ampli- 

 fier decay constant X as they are by the time constant 6 of the applied 

 signal. The magnitude of the error in any specific case is readily cal- 

 culated from Eq. (5.15). If, for example, \/9 = 50, the response meas- 

 ured at time t = SO will be in error by 36 per cent, and the response for 

 t = 56 is negative and greater in magnitude than the true signal. The 

 value of X for ^ = 1 msec, is 50 msec, in this example, and corresponds 

 to a circuit response 30 per cent below the mid band value at a fre- 

 quency of 3 cycles/sec. This response is by usual standards very good 

 indeed, but from the example is seen to give rather poor simulation of 

 the true curve. 



