Determine the Fourier series representations for the given signals in Figure P3.22(a) and (b).

For $k=0$, the coefficient $a_0$ is the average value of the signal over one period.

$$a_0 = \frac{1}{T} \int_{-T/2}^{T/2} x(t) dt = \frac{1}{2} \int_{-1}^{1} t \, dt$$ $$a_0 = \frac{1}{2} \left[ \frac{t^2}{2} \right]_{-1}^{1} = \frac{1}{2} \left( \frac{1^2}{2} - \frac{(-1)^2}{2} \right) = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{2} \right) = 0$$

$$a_k = \frac{1}{2} \int_{-1}^{1} t e^{-jk\pi t} dt$$

We use integration by parts, $\int u \, dv = uv - \int v \, du$.
Let $u = t$ and $dv = e^{-jk\pi t} dt$.
Then $du = dt$ and $v = \int e^{-jk\pi t} dt = \frac{e^{-jk\pi t}}{-jk\pi}$.

$$a_k = \frac{1}{2} \left[ t \frac{e^{-jk\pi t}}{-jk\pi} \right]_{-1}^{1} - \frac{1}{2} \int_{-1}^{1} \frac{e^{-jk\pi t}}{-jk\pi} dt$$ $$a_k = \frac{1}{2} \left[ \frac{e^{-jk\pi}}{-jk\pi} - \frac{(-1)e^{jk\pi}}{-jk\pi} \right] - \frac{1}{2(-jk\pi)} \int_{-1}^{1} e^{-jk\pi t} dt$$

We know that $e^{jk\pi} = (-1)^k$ and $e^{-jk\pi} = (-1)^k$.

$$a_k = \frac{1}{2} \left[ \frac{(-1)^k}{-jk\pi} + \frac{(-1)^k}{-jk\pi} \right] + \frac{1}{2jk\pi} \left[ \frac{e^{-jk\pi t}}{-jk\pi} \right]_{-1}^{1}$$ $$a_k = \frac{1}{2} \left[ \frac{2(-1)^k}{-jk\pi} \right] + \frac{1}{2jk\pi} \left[ \frac{e^{-jk\pi} - e^{jk\pi}}{-jk\pi} \right]$$ $$a_k = \frac{(-1)^k}{-jk\pi} + \frac{1}{2jk\pi} \left[ \frac{(-1)^k - (-1)^k}{-jk\pi} \right]$$ $$a_k = \frac{(-1)^k}{-jk\pi} + \frac{1}{2jk\pi} \left[ 0 \right]$$ $$a_k = \frac{(-1)^k}{-jk\pi} = \frac{j(-1)^k}{k\pi}$$

$$x(t) = \sum_{k=-\infty, k \neq 0}^{\infty} \frac{j(-1)^k}{k\pi} e^{jk\pi t}$$

The signal is an odd function, so $a_0 = 0$ and $a_k = -a_{-k}$.
We can also express this using sine terms:

$$x(t) = \sum_{k=1}^{\infty} \frac{2(-1)^{k+1}}{k\pi} \sin(k\pi t)$$

Part (b): The signal $x(t)$ in Figure P3.22(b) is a periodic trapezoidal wave.
From the figure, we can observe that the signal is periodic with period $T = 6$.
The fundamental angular frequency is $\omega_0 = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}$.
We can define one period of the signal from $t = -3$ to $t = 3$.
The signal is an even function, so $a_k$ will be real and $a_k = a_{-k}$.
The coefficients $b_k$ in the trigonometric Fourier series will be zero.
The function $x(t)$ over one period (e.g., from $t=-3$ to $t=3$) can be described as:

$$x(t) = \begin{cases} 0 & -3 \le t < -2 \\ t+2 & -2 \le t < -1 \\ 1 & -1 \le t < 1 \\ -t+2 & 1 \le t < 2 \\ 0 & 2 \le t \le 3 \end{cases}$$

$$a_0 = \frac{1}{T} \int_{-T/2}^{T/2} x(t) dt = \frac{1}{6} \int_{-3}^{3} x(t) dt$$ $$a_0 = \frac{1}{6} \left( \int_{-2}^{-1} (t+2) dt + \int_{-1}^{1} 1 \, dt + \int_{1}^{2} (-t+2) dt \right)$$ $$a_0 = \frac{1}{6} \left( \left[ \frac{t^2}{2} + 2t \right]_{-2}^{-1} + \left[ t \right]_{-1}^{1} + \left[ -\frac{t^2}{2} + 2t \right]_{1}^{2} \right)$$ $$a_0 = \frac{1}{6} \left( \left( \left(\frac{1}{2} - 2\right) - \left(\frac{4}{2} - 4\right) \right) + (1 - (-1)) + \left( \left(-\frac{4}{2} + 4\right) - \left(-\frac{1}{2} + 2\right) \right) \right)$$ $$a_0 = \frac{1}{6} \left( \left( -\frac{3}{2} - (-2) \right) + 2 + \left( (2) - \left(\frac{3}{2}\right) \right) \right)$$ $$a_0 = \frac{1}{6} \left( \left( -\frac{3}{2} + \frac{4}{2} \right) + 2 + \left( \frac{4}{2} - \frac{3}{2} \right) \right)$$ $$a_0 = \frac{1}{6} \left( \frac{1}{2} + 2 + \frac{1}{2} \right) = \frac{1}{6} (1 + 2) = \frac{3}{6} = \frac{1}{2}$$

Since $x(t)$ is an even function, $a_k = \frac{2}{T} \int_{0}^{T/2} x(t) \cos(k\omega_0 t) dt$.

$$a_k = \frac{2}{6} \int_{0}^{3} x(t) e^{-jk\frac{\pi}{3} t} dt = \frac{1}{3} \left( \int_{0}^{1} 1 \cdot e^{-jk\frac{\pi}{3} t} dt + \int_{1}^{2} (-t+2) e^{-jk\frac{\pi}{3} t} dt \right)$$

Alternatively, using the cosine form for even functions:

$$a_k = \frac{2}{T} \int_{0}^{T/2} x(t) \cos(k\omega_0 t) dt = \frac{2}{6} \left( \int_{0}^{1} 1 \cdot \cos(k\frac{\pi}{3} t) dt + \int_{1}^{2} (-t+2) \cos(k\frac{\pi}{3} t) dt \right)$$ $$a_k = \frac{1}{3} \left( \left[ \frac{\sin(k\frac{\pi}{3} t)}{k\frac{\pi}{3}} \right]_{0}^{1} + \int_{1}^{2} (-t+2) \cos(k\frac{\pi}{3} t) dt \right)$$ $$a_k = \frac{1}{3} \left( \frac{\sin(k\frac{\pi}{3})}{k\frac{\pi}{3}} + \int_{1}^{2} (-t+2) \cos(k\frac{\pi}{3} t) dt \right)$$

For the integral $\int (-t+2) \cos(k\frac{\pi}{3} t) dt$, use integration by parts:
Let $u = -t+2$, $dv = \cos(k\frac{\pi}{3} t) dt$.
Then $du = -dt$, $v = \frac{\sin(k\frac{\pi}{3} t)}{k\frac{\pi}{3}}$.

$$\int_{1}^{2} (-t+2) \cos(k\frac{\pi}{3} t) dt = \left[ (-t+2) \frac{\sin(k\frac{\pi}{3} t)}{k\frac{\pi}{3}} \right]_{1}^{2} - \int_{1}^{2} \frac{\sin(k\frac{\pi}{3} t)}{k\frac{\pi}{3}} (-dt)$$ $$= \left[ (-t+2) \frac{\sin(k\frac{\pi}{3} t)}{k\frac{\pi}{3}} \right]_{1}^{2} + \frac{1}{k\frac{\pi}{3}} \int_{1}^{2} \sin(k\frac{\pi}{3} t) dt$$ $$= \left( (0) - (1) \frac{\sin(k\frac{\pi}{3})}{k\frac{\pi}{3}} \right) + \frac{1}{k\frac{\pi}{3}} \left[ -\frac{\cos(k\frac{\pi}{3} t)}{k\frac{\pi}{3}} \right]_{1}^{2}$$ $$= -\frac{\sin(k\frac{\pi}{3})}{k\frac{\pi}{3}} - \frac{1}{(k\frac{\pi}{3})^2} (\cos(k\frac{2\pi}{3}) - \cos(k\frac{\pi}{3}))$$

Substitute this back into the expression for $a_k$:

$$a_k = \frac{1}{3} \left( \frac{\sin(k\frac{\pi}{3})}{k\frac{\pi}{3}} - \frac{\sin(k\frac{\pi}{3})}{k\frac{\pi}{3}} - \frac{1}{(k\frac{\pi}{3})^2} (\cos(k\frac{2\pi}{3}) - \cos(k\frac{\pi}{3})) \right)$$ $$a_k = -\frac{1}{3} \frac{1}{(k\frac{\pi}{3})^2} (\cos(k\frac{2\pi}{3}) - \cos(k\frac{\pi}{3}))$$ $$a_k = -\frac{3}{(k\pi)^2} (\cos(k\frac{2\pi}{3}) - \cos(k\frac{\pi}{3}))$$

$$x(t) = a_0 + \sum_{k=1}^{\infty} a_k \cos(k\omega_0 t)$$ $$x(t) = \frac{1}{2} + \sum_{k=1}^{\infty} -\frac{3}{(k\pi)^2} (\cos(k\frac{2\pi}{3}) - \cos(k\frac{\pi}{3})) \cos(k\frac{\pi}{3} t)$$

Final Answer:

a) The Fourier series representation for $x(t)$ in Figure P3.22(a) is:

$$x(t) = \boxed{\sum_{k=-\infty, k \neq 0}^{\infty} \frac{j(-1)^k}{k\pi} e^{jk\pi t}}$$

or equivalently:

$$x(t) = \boxed{\sum_{k=1}^{\infty} \frac{2(-1)^{k+1}}{k\pi} \sin(k\pi t)}$$

b) The Fourier series representation for $x(t)$ in Figure P3.22(b) is:

$$x(t) = \boxed{\frac{1}{2} + \sum_{k=1}^{\infty} -\frac{3}{(k\pi)^2} \left(\cos\left(k\frac{2\pi}{3}\right) - \cos\left(k\frac{\pi}{3}\right)\right) \cos\left(k\frac{\pi}{3} t\right)}$$