What will occur when a non-linear amplifier is used with a single-sideband phone transmitter?

Distortion.

From kj6prf:

Linear amplifiers preserve the shape of the input signal, but nonlinear amplifiers do not. Thus when a nonlinear amplifier is used, the waveform's shape changes, and this is called distortion.

Please see the Resources System Analysis Cadence site for the article Amplifier Non-Linearity Leads to Output Signal Distortions

See Wikipedia's article Single-sideband modulation

Also, see the Electronics Notes site for the article Single Sideband Modulation, SSB

Last edited by markadlerdallas. Register to edit

Tags: none

To produce a single-sideband suppressed carrier transmission it is necessary to ____ the carrier and to ____ the unwanted sideband.

Cancel, filter.

To produce a single-sideband suppressed carrier transmission it is necessary to CANCEL the carrier and to FILTER the unwanted sideband.

Please see the Resources System Analysis Cadence site for the article Amplifier Non-Linearity Leads to Output Signal Distortions

See Wikipedia's article Single-sideband modulation

Also, see the Electronics Notes site for the article Single Sideband Modulation, SSB

See Ham Radio School for the article Understanding Single Sideband (SSB)

And, see the Sgcworld site for the article What is Single Side-band?

Last edited by markadlerdallas. Register to edit

Tags: none

In a single-sideband phone signal, what determines the PEP-to-average power ratio?

The speech characteristics.

Which means what is the input into the SSB.

See the Apps Dtic Mil site for the pdf document from the US Department of Agriculture, titled Technical Report ERL 131-ITS 92 Required Signal-to-Noise Ratios for HF Communication Systems

Also, see the k0bg site for the article on Audio Transmit

And, see the CPII site for the pdf document titled LINEAR AMPLIFIER AND SINGLE SIDEBAND SERVICE

For more information, please see the Eng Auburn Edu site for the pdf article SSB GENERATION - THE PHASING METHOD

See Wikipedia's article Peak envelope power

Last edited by markadlerdallas. Register to edit

Tags: none

What is the approximate ratio of peak envelope power to average power during normal voice modulation peak in a single-sideband phone signal?

2.5 to 1.

From meetar:

A modulation envelope shows how a carrier wave's amplitude changes over time, the envelope power is the power value of that envelope, and the peak envelope power is the highest power value within some time frame.

However, if a carrier is unmodulated, its power isn't changing, which means its modulation envelope is flat. And as with any unchanging value, its peak equals its average, and the ratio of two equal values is
\[2.5:1 \]

(Note that as power varies directly with amplitude\(^2\), it's convenient to think of the envelope as representing both.)

(Note also that the question asks about the ratio of two power values, not the values themselves, so RMS calculations aren't involved here.)

See the RFG Global Net site for the article Basics Of Power Measurement — Average Or Peak?

Also, please see the Resources PCB Cadence site for the Peak Envelope Power and Your RF Signal Chain Simulations

Last edited by markadlerdallas. Register to edit

Tags: none

What is the output peak envelope power from a transmitter as measured on an oscilloscope showing 200 volts peak-to-peak across a 50-ohm load resistor?

100 watts.

From wp2ahg:

Start by converting peak-to-peak voltage to RMS:

200 volts = 100 volts PEP x 0.707 = 70.7 volts RMS

Now, use the Ohm's Law formula P = E² / R
P = 70.7 volts x 70.7 volts / 50 ohms
P = 4998.49 volts / 50 ohms
P = 99.97 watts

100 watts is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50-ohm resistor connected to the transmitter output.

Use the combined power and Ohm's law equation:

\[P = \frac{E^2}{R}\]

where: \begin{align} P &= \text{Peak Envelope Power}\\ E &= \text{Root Mean Square (RMS) Voltage} \end{align}

First find the \(E_{RMS}\):

\[E_{RMS} = \frac{\text{Peak Envelope Voltage}}{\sqrt{2}}\]

To find the Peak Envelope Voltage (PEV): \begin{align} \text{PEV} &= \frac{\text{Peak-to-Peak Voltage}}{2}\\ &= \frac{200\ \text{V}}{2}\\ &= 100\ \text{V} \end{align}

So now the Peak Envelope Power (PEP) output in this case can be calculated as:

\begin{align} P_{PEP} &= \frac{(E_{RMS})^2}{R}\\ &= \frac{ \Big ( \frac{ 100\ \text{V}}{ \sqrt{2} } \Big ) ^2}{50\ \Omega}\\ &= 100\ \text{W} \end{align}

Last edited by markadlerdallas. Register to edit

Tags: none

What would be the voltage across a 50-ohm dummy load dissipating 1,200 watts?

245 volts.

From wscody:

The RMS voltage across a 50-ohm dummy load dissipating 1200 watts is 245 volts.

You are given:
\begin{align} Power (P) &= 1200\text{ watts}\\ Resistance (R) &= 50\ \Omega \end{align}

Combining the power circle equation \(P = I \times E\) and the Ohm's Law equation \(I = \frac{E}{R}\) gives the equation: \[P = \frac{ E^2 }{ R\ \ }\]

Rearranging to solve for \(E\):

\[E = \sqrt{ P \times R }\]

Using the values in this question:

\begin{align} E &= \sqrt{ 1200\text{ W} \times 50\ \Omega }\\ &= \sqrt{ 60000 }\text{ V}\\ &= 100\sqrt{ 6 }\text{ V}\\ &= 245\text{ V} \end{align}

Alternate Hint: \(E= \sqrt{P \times R}\). \(1200 \times 50 = 60,000\); (\(\sqrt{60,000} = 245\))

Silly Hint: The correct answer is the only one that has both a 2 (as in 1200 watts) and a 5 (as in 50 ohms).

Last edited by markadlerdallas. Register to edit

Tags: none