All energy signals will have an average Power of
zero
In the study of signals and systems, signals are often classified based on their energy and power properties. Two fundamental classifications are energy signals and power signals.
An energy signal is a signal that has finite total energy over its entire duration. This means that the integral (for continuous-time signals) or summation (for discrete-time signals) of the squared magnitude of the signal over all time is a finite, non-zero value.
Mathematically, for a continuous-time signal \(x(t)\), the total energy \(E\) is defined as:
\( E = \int_{-\infty}^{\infty} |x(t)|^2 dt \)
For an energy signal, the condition is \(0 < E < \infty\).
Similarly, for a discrete-time signal \(x[n]\), the total energy \(E\) is defined as:
\( E = \sum_{n=-\infty}^{\infty} |x[n]|^2 \)
Again, for a discrete-time energy signal, \(0 < E < \infty\).
Energy signals typically "die out" or approach zero as time \(|t|\) or \(|n|\) goes to infinity. Examples include decaying exponentials or finite-duration pulses.
The average power of a signal represents the average rate at which the signal delivers energy over an infinite time interval. It is calculated by taking the total energy over a large time interval and dividing it by the duration of that interval, then taking the limit as the interval approaches infinity.
For a continuous-time signal \(x(t)\), the average power \(P_{avg}\) is defined as:
\( P_{avg} = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt \)
For a discrete-time signal \(x[n]\), the average power \(P_{avg}\) is defined as:
\( P_{avg} = \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^{N} |x[n]|^2 \)
Signals that have finite non-zero average power are called power signals. Examples include periodic signals or random signals that persist over time.
Now let's consider the relationship between the finite energy of an energy signal and its average power. For an energy signal \(x(t)\), the total energy \(E = \int_{-\infty}^{\infty} |x(t)|^2 dt\) is finite. The average power formula involves integrating over a finite interval \([-T, T]\) and then dividing by the interval duration \(2T\) as \(T \to \infty\).
The integral \(\int_{-T}^{T} |x(t)|^2 dt\) represents the energy of the signal within the interval \([-T, T]\). As \(T \to \infty\), this integral approaches the total energy \(E\). So, for an energy signal:
\( P_{avg} = \lim_{T \to \infty} \frac{1}{2T} \left( \int_{-T}^{T} |x(t)|^2 dt \right) \)
Since \( \lim_{T \to \infty} \int_{-T}^{T} |x(t)|^2 dt = E \) (which is a finite value for an energy signal), the formula becomes:
\( P_{avg} = \lim_{T \to \infty} \frac{1}{2T} \times E \)
As \(T\) approaches infinity, the term \(\frac{1}{2T}\) approaches zero. Since \(E\) is a finite constant, the product \(\frac{1}{2T} \times E\) approaches \(0 \times E = 0\).
The same logic applies to discrete-time energy signals. The summation \(\sum_{n=-N}^{N} |x[n]|^2\) approaches the total finite energy \(E\) as \(N \to \infty\). The average power is then \(\lim_{N \to \infty} \frac{1}{2N+1} \times E\). As \(N \to \infty\), \(\frac{1}{2N+1} \to 0\), leading to \(P_{avg} = 0 \times E = 0\).
Therefore, any signal that has finite total energy over all time must have an average power of zero when averaged over infinite time.
Signals are categorized based on whether their energy or power is finite and non-zero over infinite time:
| Signal Type | Total Energy (E) | Average Power (Pavg) | Example |
|---|---|---|---|
| Energy Signal | Finite (\(0 < E < \infty\)) | Zero (\(P_{avg} = 0\)) | Decaying exponential, finite pulse |
| Power Signal | Infinite (\(E = \infty\)) | Finite Non-Zero (\(0 < P_{avg} < \infty\)) | Sinusoid, unit step function |
| Neither | Infinite (\(E = \infty\)) | Infinite (\(P_{avg} = \infty\)) | Ramp function (\(x(t) = t u(t)\)) |
Since the question specifically asks about energy signals, and we have established that energy signals have finite energy \(E\) and thus zero average power \(P_{avg}\), the correct answer is zero.
| Concept | Description | Condition for Energy Signal | Condition for Power Signal |
|---|---|---|---|
| Total Energy (E) | Sum/integral of squared magnitude over all time. | \(0 < E < \infty\) | \(E = \infty\) |
| Average Power (Pavg) | Average rate of energy over infinite time. | \(P_{avg} = 0\) | \(0 < P_{avg} < \infty\) |
It's important to note that a signal cannot be both an energy signal and a power signal simultaneously, because for an energy signal, \(E\) is finite and \(P_{avg}\) is zero, while for a power signal, \(E\) is infinite and \(P_{avg}\) is finite and non-zero. These classifications cover most signals encountered in typical signal processing and communication systems.
The classification helps in choosing appropriate analysis tools. For example, the Fourier Transform is more directly applicable to energy signals, while the Power Spectral Density (PSD) is used for analyzing power signals.
Signals that grow unboundedly in magnitude over time (like \(e^t\) or \(t u(t)\)) often have both infinite energy and infinite average power, and they don't fit neatly into either the energy or power signal category as strictly defined with finite non-zero values.
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A. Type N system
B. Type 3 system
C. Type 2 system
D. Type 4 system
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Statement I: The scaling theorem relates scale changes in frequency domain to the consequent changes in scale in the time domain.
Statement II: The scaling theorem is not applicable to an impulse signal.
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