AnaCom (Chapter 2) (Mid)

Answer:

2.1-2(a)

We have two signals $x(t)$ and $y(t)$ in Fig. P2.1-2a (not fully drawn here, but let's assume they are simple pulses).

Step 1: Energy of a signal

The energy of a continuous-time signal $x(t)$ is:

$$E_x={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\mid }x(t){\mathrm{\mid }}^{2}dt$$

Similarly for $y(t)$:

$$E_y={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\mid }y(t){\mathrm{\mid }}^{2}dt$$

Step 2: Energy of $x(t)+y(t)$ and $x(t)-y(t)$

Let’s compute $E_{x+y}$:

$$E_{x+y}=\int [x(t)+y(t){]}^{2}dt$$$$=\int [x(t{)}^2+y(t{)}^2+2x(t)y(t)]dt$$$$=E_x+E_y+2\int x(t)y(t)dt$$

Similarly:

$$E_{x-y}=E_x+E_y-2\int x(t)y(t)dt$$

Step 3: When are they equal to $E_x+E_y$?

If $\int x(t)y(t)dt=0$ (i.e., $x$ and $y$ are orthogonal), then:

$$E_{x+y}=E_x+E_y$$$$E_{x-y}=E_x+E_y$$

In part (a) of the figure (assumed shape), $x(t)$ and $y(t)$ are non-overlapping in time or arranged so that $\int x(t)y(t)dt=0$.
So yes, $E_{x+y}=E_{x-y}=E_x+E_y$.

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2.1-2(b)

Repeat for the second signal pair in Fig. P2.1-2b.
If $x(t)$ and $y(t)$ are orthogonal, same result.
If not orthogonal, then $E_{x+y}\mathrm{\ne }{E}_{x-y}$.

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2.1-3

Power of a sinusoid $C\cos ({\omega }_{0}t+\theta )$:

For a sinusoid, average power = $\frac{C^2}{2}$.

Reason:

$${\cos }^2({\omega }_{0}t+\theta )=\frac{1+\cos (2{\omega }_{0}t+2\theta )}{2}$$

Average over one period:
$⟨{\cos }^2(\cdot )⟩=\frac{1}{2}$
So:

$$P=\frac{C^2}{2}$$

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2.1-4

Given: $g(t)=C_1\cos ({\omega }_{1}t+{\theta }_1)+C_2\cos ({\omega }_{2}t+{\theta }_2)$, with ${\omega }_1={\omega }_2=\omega$.

Let’s find power:

We know:
Power of $A\cos (\omega t+\varphi )$ is $A^2\mathrm{/}2$.
But here the two cosines have same frequency, so they add phasorially:

$$g(t)=\text{Re}[C_1{e}^{j(\omega t+{\theta }_1)}+C_2{e}^{j(\omega t+{\theta }_2)}]$$$$=\text{Re}[e^{j\omega t}(C_1{e}^{j{\theta }_1}+C_2{e}^{j{\theta }_2})]$$

Let $C{e}^{j\theta }=C_1{e}^{j{\theta }_1}+C_2{e}^{j{\theta }_2}$.
Then $C^2=C_1^2+C_2^2+2C_1{C}_2\cos ({\theta }_1-{\theta }_2)$.

So $g(t)=C\cos (\omega t+\theta )$, whose power is $C^2\mathrm{/}2$:

$$P=\frac{C_1^2+C_2^2+2C_1{C}_2\cos ({\theta }_1-{\theta }_2)}{2}$$

Not equal to $(C_1^2+C_2^2)\mathrm{/}2$ unless $\cos ({\theta }_1-{\theta }_2)=0$.

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2.1-5

Given a periodic $g(t)$ (triangular or rectangular waveform in Fig. P2.1-5).
Let’s assume it’s a square wave amplitude $A$, period $T$, duty cycle 50%.

Power of periodic signal:

$$P=\frac{1}{T}{\int }_0^T[g(t){]}^{2}dt$$

For square wave from $-A$ to $A$, half time $A$, half time $-A$:

$$g(t{)}^2=A^2$$

So $P=A^2$.

(a) $-g(t)$

Power = $A^2$ (same, squaring removes sign).

(b) $2g(t)$

Power = $(2A{)}^2=4A^2$.

(c) $cg(t)$

Power = $c^2{A}^2$.

RMS value = $\sqrt{\text{Power}}$.

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2.1-6

Find power and RMS for given figures.

(a) Fig. P2.1-6a

If it’s a constant $A$, then:

$$P=A^2,\text{RMS}=\mathrm{\mid }A\mathrm{\mid }$$

(b) Fig. 2.16

If it’s a sine amplitude $A$:

$$P=A^2\mathrm{/}2,\text{RMS}=A\mathrm{/}\sqrt{2}$$

(c) Fig. P2.1-6b

If it’s a shifted sine, same power $A^2\mathrm{/}2$.

(d) Fig. P2.7-4a

If it’s a periodic triangular wave amplitude $A$:
Let’s compute:
For triangle wave from 0 to $A$ linearly, then down:
Average of square = $A^2\mathrm{/}3$ for full triangle centered at zero? Actually, for a triangle wave peak $A$, $P=A^2\mathrm{/}3$.

(e) Fig. P2.7-4c

If it’s a full-wave rectified sine amplitude $A$:
$g(t)=A\mathrm{\mid }\sin (\omega t)\mathrm{\mid }$
Square: $A^2{\sin }^2(\omega t)$ averaged = $A^2\mathrm{/}2$? Wait, check:
${\sin }^2$ average = 1/2, so $P=A^2\mathrm{/}2$ for full-wave rectified sine.

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Final summary:

• Energy signals: use integration of square.

• Power of sinusoid: $C^2\mathrm{/}2$.

• Power of sum of equal-frequency sinusoids: depends on phase difference.

• RMS = $\sqrt{\text{Power}}$.