**Question #1 Convert the following numbers (ratios) into decibels **

All answers in dB must be written with one decimal place – ONLY.

** **

- 2.0 =
- 200.0 =
- 0.02 =
- 10
^{6 =} - 4 ´ 10
^{-4}=

**Question #2 Convert the following values into decibels with the correct units **

** **

**Units must be correct**

** **

- 2.0 watts =
- 2000 hertz =
- 500 kelvins =
- 10
^{7 }hertz = - 40 mW =

**Question #3 Convert the following decibel quantities into the corresponding linear values**

(2 points)

** **

- 3.0 dBW =
- -20 dBW =
- -30 dBm =
- 30 dBK =
- 60.0 dBHz =

** **

**Question #4 Calculate the following expressions using dB throughout **

(Do not calculate the answer in linear arithmetic and then convert to dB – show your working.)

** **

- 4.0 ´ 8.0 = 6.0 + 9.0 =
- 200 / 8.0 = 23.0 – 9.0 =
- 10
^{6}/ 250 = 60.0 – 24.0 = - 1.38 ´ 10
^{-23}W/K/Hz ´ 10^{6 K}´ 1000 Hz = -228.6 + 60 + 30 =

(Noise calculation for N = k T B)

- Do this calculation in dB units, give the result in dB units:

[ 4 × p × 36,500 × 10^{3} / 0.05 ]^{2} =

**Question #5**

Earlier in this exercise you were told to round dB units to nearest 0.1 dB. Decibels were originally applied to sound pressure levels and it is said that a 1.0 dB difference is the smallest difference that most folks can hear. (The reference level is 1 dyne/cm^{2} so a sound pressure level of 1 dyne/cm^{2} is 0 dB.) Determine the % difference between 0 dB (ratio of 1) and 1 dB, 0 dB and 0.1 dB, and 0 dB and 0.01 dB. Hint: convert the dB values back to a ratios and then determine the % difference in the two values.

**Question #6**

Remember that dB is ALWAYS a ratio two powers. Yet you often see dB = 20 log (V_{2}/V_{1}), the ratio of two voltages. It is WRONG but every communications engineer I know (including myself) does it. Consider dB = 10 log (P_{2}/P_{1}) = 10 log (V_{2}^{2}/R_{2})/ (V_{1}^{2}/R_{1}). Expand this so dB = 20 log (V_{2}/V_{1}) +(?). Explain when dB = 20 log (V_{2}/V_{1}) is correct and when it is incorrect.

** **

**Question #7 Multiple choice: mark (underline, bold, change color, etc) ALL answers that apply **

** **

** **A signal in the IF section of an FM radio receiver occupies the frequency band 10.61MHz to 10.79 MHz. The bandwidth of this signal is

180 MHz 180 kHz 180,000 Hz 10.7 MHz

**Question #8 Multiple choice: circle ALL answers that apply**

** **

** **A signal in the IF section of an FM radio receiver occupies the frequency band 10.61MHz to 10.79 MHz. There is another FM signal occupying the frequency range 11.01 MHz to 11.19 MHz. A filter is used to extract the first signal. The type of filter needed is

Low pass High pass Band pass Band stop

** **

**Question #9 Multiple choice: mark (underline, bold, change color, etc) ALL answers that apply **

At frequencies well above the 3 dB frequency of a fourth order Butterworth low pass filter, the attenuation of the filter increases at a rate of

80 dB per octave 80 dB per decade 36 dB per decade 24 dB per octave

**Question #8 Write a single paragraph on each of these topics **

** **

** **Explain why a *hybrid transformer* is needed in a telephone handset. What is the relationship between the hybrid and *sidetone*

Where else in a telephone system would you find a hybrid transformer? What purpose does it serve?