CORE1240 HW2

Q1

(1)

(2)

The bandwidth of $S1$ is $10\mathrm{kHz}$.
The bandwidth of $S2$ is $10\mathrm{kHz}$.

(3)

We can apply a band-pass filter, which centers at $32\mathrm{MHz}$ with a bandwidth of $10\mathrm{kHz}$, in order words, it has a low cutoff frequency of $31995\mathrm{kHz}$ and high cutoff frequency of $32005\mathrm{kHz}$.

(4)

1. Multiply the modulated signal by the carrier frequency.
2. Apply a low-pass filter with cutoff frequency $5\mathrm{kHz}$.

(5)

Q2

(1)

(2)

1111000110

(3)

3 bits are in error.

Q3

(1)

$$Y=\bar{A}\bar{B}\bar{C}+\bar{A}B\bar{C}$$

(2)

Q4

(1)

The biggest binary number is 111111, which is 63 in decimal value.

(2)

$$2^n-1\mathrm{bits}$$

(3)

$$\begin{array}{rl}{2}^n& \ge 17\\ n& \ge \frac{\log 17}{\log 2}\\ n& \ge 4.09\end{array}$$

The minimum number of bits required is 5 bits.

(4)

The binary representation of 238 is 11101110