Therefore,
The transient response and its Fourier spectrum are shown in Figure 1.6.
Figure 1.6 Time and frequency domain representations of the transient response of an R–C series circuit.
Example 1.3
The impulse response of a dynamic system is its output in response to a brief input pulse signal, called an impulse. The impulse response of the damped vibration of a one‐degree‐of‐freedom mass‐spring system of mass M, stiffness K, and coefficient of damping R (see Chapter 2 of this book) is given by
where A = (Mωd)−1, α = R/2 M and λ = ωd is known as the damped “natural” angular frequency. Find the Fourier spectrum representation of this impulse response.
Solution
Using the mathematical property ejθ = cos θ + j sin θ, we can write
The impulse response and its Fourier spectrum are shown in Figure 1.7. We notice that replacing α and λ by the corresponding values in terms of the stiffness K, mass M, and damping constant R, of the damped mass‐spring system, the Fourier spectrum becomes (compare with Eq. (2.18))
Figure 1.7 Time and frequency domain representations of the transient response of the impulse response of a damped vibration of a mass‐spring system.
1.3.4 Mean Square Values
In the case of the pure tone a useful quantity to determine is the mean square value, i.e. the time average of the signal squared 〈x2(t)〉t [8]
where 〈〉t denotes a time average.
For the pure tone in Figure 1.2a then we obtain
where A is the signal amplitude.
The root mean square value is given by the square root of 〈x2(t)〉t or
For the general case of the complex pure tone in Eq. (1.1) or (1.2) we obtain:
or
(1.11)
since
Example 1.4
Determine the mean square and rms values of the signal in Figure 1.3.
Solution
We can use Eq. (1.7) to determine its mean square value,
The same result is obtained from its Fourier series representation using Eq. (1.10):
Recalling that the root mean square value is given by the square root of the mean square value, the rms value of this saw tooth signal is
1.3.5 Energy and Power Spectral Densities
