From 876c00011e0f7efc52bcc7c20e2a03bf04f7f63d Mon Sep 17 00:00:00 2001 From: eneller Date: Tue, 7 Apr 2026 01:30:59 +0200 Subject: [PATCH] a5/a6 --- a5.md | 19 +++++++++++++++++++ a6.md | 3 +++ 2 files changed, 22 insertions(+) create mode 100644 a5.md create mode 100644 a6.md diff --git a/a5.md b/a5.md new file mode 100644 index 0000000..f739b5e --- /dev/null +++ b/a5.md @@ -0,0 +1,19 @@ +# Suppressed Carrier Wave +Derive expressions from which it will be evident whether the +communication channel has more power as an AM signal with +a suppressed carrier wave or as an AM signal with the carrier +wave present. Assume the voltage signal, $v(t)$ and $m(t)$, as in +the previous figure. Modulating (LF) sinusoidal signal has +amplitude $A$ and frequency $f_{LF}$ Hz, cosine wave RF carrier has +amplitude $1$ and frequency $f_{VF}$. + + +According to $P_t = P_c + P_{USB} + P_{LSB} $ the total transmitting power has to be split among carrier, upper and lower side band. +With the carrier wave suppressed, the modulating signal will take up all of the available transmission power instead. + +For a standard AM signal with the above requirements, the formula is: +$$ s(t) = (1+A \cos(2\pi f_{LF}t)) \cdot \cos(2\pi f_{VF} t)$$ + +With the power of a wave signal given by $P=\frac{(v_m/\sqrt{2})^2}{2}$ (where $v_m$ is max of cosine signal), +our sideband power is $2* \frac{(A/\sqrt{2})^2}{2} = \frac{A^2}{2}$ +and our carrier power $\frac{1}{2}$, equal to the power gained in DSB-SC. diff --git a/a6.md b/a6.md new file mode 100644 index 0000000..d0a14dc --- /dev/null +++ b/a6.md @@ -0,0 +1,3 @@ +# Time Synchronization +Time sync between sender and receiver is important because the carrier wave has to be used to demodulate the signal in the receiver. +If there is a time shift, there will be a phase error introduced by the demodulation, resulting in a distorted signal. \ No newline at end of file