E1C02 09/14/2010
13:35:17 Page 49
where I represents the time-dependent current in amperes. Establish the mean and rms values of
current over a time from 0 to t f , with t f ¼ p and then with t f ¼ 2p. How do the results relate to the
power dissipated in the resistor?
KNOWN I(t) ¼ 10 sin t
FIND I and I rms with t f ¼ p and 2p
SOLUTION The average value for a time from 0 to t f is found from Equation 2.1 as
I ¼
ð t f
0
I t
ð Þdt
ð t f
0
dt
¼
ð t f
0
10 sin tdt
t f
Evaluation of this integral yields
I ¼
1
t f
À10 cos t
½
t f
0
With t f ¼ p, the average value, I, is 20=p. For t f ¼ 2p, the evaluation of the integral yields an
average value of zero.
The rms value for the time period 0 to t f is given by the application of Equation 2.4, which yields
I rms ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1
t f
ð t f
0
I t
ð Þ
2 dt
s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1
t f
ð t f
0
10 sin t
ð
Þ
2 dt
s
This integral is evaluated as
I rms ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
100
t f
À
1
2
cos t sin t þ
t
2
t f
0
s
For t f ¼ p, the rms value is
ffiffiffiffiffi
50
p
. Evaluation of the integral for the rms value with t f ¼ 2p also
yields
ffiffiffiffiffi
50
p
.
COMMENT Although the average value over the period 2p is zero, the power dissipated in the
resistor must be the same over both the positive and negative half-cycles of the sine function. Thus,
the rms value of current is the same for the time period of p and 2p and is indicative of the rate at
which energy is dissipated.
2.4 SIGNAL AMPLITUDE AND FREQUENCY
A key factor in measurement system behavior is the nature of the input signal to the system. A means is
needed to classify waveforms for both the input signal and the resulting output signal relative to their
magnitude and frequency. It would be very helpful if the behavior of measurement systems could be
defined in terms of their response to a limited number and type of input signals. This is, in fact, exactly
the case. Avery complex signal, even one that is nondeterministic in nature, can be approximated as an
infinite series of sine and cosine functions, as suggested in Table 2.1. The method of expressing such a
complex signal as a series of sines and cosines is called Fourier analysis.
2.4 Signal Amplitude And Frequency 49
13:35:17 Page 49
where I represents the time-dependent current in amperes. Establish the mean and rms values of
current over a time from 0 to t f , with t f ¼ p and then with t f ¼ 2p. How do the results relate to the
power dissipated in the resistor?
KNOWN I(t) ¼ 10 sin t
FIND I and I rms with t f ¼ p and 2p
SOLUTION The average value for a time from 0 to t f is found from Equation 2.1 as
I ¼
ð t f
0
I t
ð Þdt
ð t f
0
dt
¼
ð t f
0
10 sin tdt
t f
Evaluation of this integral yields
I ¼
1
t f
À10 cos t
½
t f
0
With t f ¼ p, the average value, I, is 20=p. For t f ¼ 2p, the evaluation of the integral yields an
average value of zero.
The rms value for the time period 0 to t f is given by the application of Equation 2.4, which yields
I rms ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1
t f
ð t f
0
I t
ð Þ
2 dt
s
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1
t f
ð t f
0
10 sin t
ð
Þ
2 dt
s
This integral is evaluated as
I rms ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
100
t f
À
1
2
cos t sin t þ
t
2
t f
0
s
For t f ¼ p, the rms value is
ffiffiffiffiffi
50
p
. Evaluation of the integral for the rms value with t f ¼ 2p also
yields
ffiffiffiffiffi
50
p
.
COMMENT Although the average value over the period 2p is zero, the power dissipated in the
resistor must be the same over both the positive and negative half-cycles of the sine function. Thus,
the rms value of current is the same for the time period of p and 2p and is indicative of the rate at
which energy is dissipated.
2.4 SIGNAL AMPLITUDE AND FREQUENCY
A key factor in measurement system behavior is the nature of the input signal to the system. A means is
needed to classify waveforms for both the input signal and the resulting output signal relative to their
magnitude and frequency. It would be very helpful if the behavior of measurement systems could be
defined in terms of their response to a limited number and type of input signals. This is, in fact, exactly
the case. Avery complex signal, even one that is nondeterministic in nature, can be approximated as an
infinite series of sine and cosine functions, as suggested in Table 2.1. The method of expressing such a
complex signal as a series of sines and cosines is called Fourier analysis.
2.4 Signal Amplitude And Frequency 49