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Calculate the power out put of an amplifier that has an input of 15 μw and a gain of 25dB?
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 micro Watts x 316.2 = (4700)/(10^6) = 4.7 milli watts If you want to work in dB, then convert 15 micro watts to dB: 10 * log |P| = dB = 10*log |15 x 10^6| = -48.2dB ***When you have very small (ie negative) dB, it is often referred to in dBm, or 1/1000 of dB ( 30 dBm = 0 dB) so the output is -18.2dBm + 25 = 6.8dBm, or -23.2dB
Milliwatts to dbm conversion?
The 'm' in dBm means the power is referenced to 1mW. So, the power in dBm equals 10 times the log of the power in mW, or P(dBm) = 10*log(P(mW)/1mW). For example, 1W = 1000mW, so 10*log(1000/1) = 30dBm.
Which code i need to delete middle digit in 3 digit number in c plus plus?
If you know that the number input will always be three digits: output = 10 * (int)(input / 100) + (input % 10); If you want to idiot proof it (eg. too many digits): output = 10 * (int)((input % 1000) / 100) + (input % 10);
What is the pin diagram of Integrated Chip L293D?
L293D is having 20 pin IC and also 16 pin IC. description of 20 pin is: 1-enable 1 2- input 1 3- output 1 4,5,6,7,14,15,16,17- ground 8- output 2 9- input 2 10,20-vs 11-enable 2 12- input 3 13-output 3 18-output 4 19-input 4 description for 18 pin: 1-enable 1 2- input 1 3- output 1 4,5,12,13- ground 6-output 2 7- input 2 8,18-vs 9-enable 2 10-input 3 11-output 3 14-output 4 15-input 4
How can IC 74190 will give output freq as 50Hz?
The 74190 is an up/down decade counter. Counters use frequency division to achieve a counting sequence. To answer your question, it depends on the input frequency. The Qa output will divide the clock input by 2 so if the input is 100Hz, Qa's output is 50Hz. Since this is a decade (0 to 9, or truncated sequence) counter and not a binary (0 to 15, or full sequence) counter, the outputs Qb, Qc and Qd divide the input but their outputs are not symmetrical (equal time high and time low). Qb and Qc produce 2 pulses for every 10 input pulses, therefore divide the input clock by 5. Qd produces one output pulse for every 10 input pulses, therefore divides the input by 10. The easiest way to visualize this is to write out the binary count in column format, starting at 0000 and ending at 1001, and looking at each of the output patterns. To produce the 50Hz output, assuming you are not concerned over symmetry: -input clock 100Hz for 50 Hz on Qa -input clock 250Hz for Qb or Qc output of 50Hz -input clock 500Hz for Qd output of 50Hz
How do you calculate the output W of a cable with 2W input at 13dB?
13 dB less than 2 W is 0.1 W(or -10 dBw, or +20 dBm)
How do you get power output with a given power input of 20dbm and three power gain which is Ap1 equals 10 ap2 equals 4 ap3 equals 23?
Two ways to do it. In this particular problem, it's a matter of opinionwhich one is easier and which one is harder.Way #1:Convert dBm to watts, multiply by gains, convert output watts to dBm.+20 dBm = 0.1 watt.Output power = (0.1 watt) x (ap1) x (ap2) x (ap3) = 0.1 x 10 x 4 x 23 = 92 watts = +49.64 dBmWay #2:Convert each gain ratio to dB, then add all dB to input power.ap1 = 10 = 10 dBap2 = 4 = 6.02 dBap3 = 23 = 13.62 dB+20 dBm + 10dB + 6.02 dB + 13.62 dB = +49.64 dBm
Calculate the power out put of an amplifier that has an input of 15 μw and a gain of 25dB?
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 micro Watts x 316.2 = (4700)/(10^6) = 4.7 milli watts If you want to work in dB, then convert 15 micro watts to dB: 10 * log |P| = dB = 10*log |15 x 10^6| = -48.2dB ***When you have very small (ie negative) dB, it is often referred to in dBm, or 1/1000 of dB ( 30 dBm = 0 dB) so the output is -18.2dBm + 25 = 6.8dBm, or -23.2dB
A receiver requires 10nW as input power if all the system losses add up to 50dB then how much power is required from the source?
here, the power required by the receiver is the output power and that required from the source is input power. Gain in dB=10 log(output power/input power) we have, loss in dB = -gain in dB = 10 log(input power/output power) or, 50 = 10 log(input power/10nW) or, anti-log(5) = input power/10 nW so the power required from the source is antilog(5)*10nW = 1 mW
Milliwatts to dbm conversion?
The 'm' in dBm means the power is referenced to 1mW. So, the power in dBm equals 10 times the log of the power in mW, or P(dBm) = 10*log(P(mW)/1mW). For example, 1W = 1000mW, so 10*log(1000/1) = 30dBm.
What is the output number if the input number is 100?
50
Which code i need to delete middle digit in 3 digit number in c plus plus?
If you know that the number input will always be three digits: output = 10 * (int)(input / 100) + (input % 10); If you want to idiot proof it (eg. too many digits): output = 10 * (int)((input % 1000) / 100) + (input % 10);
What is the math definition for input output tables?
A table in which you put in a number and out comes another number. Usually more than one groups of numbers. And almost ALWAYS follows a rule such as: Input x3=Output or Input -23= Output Input | Output 2 | 4 10 | 20 16 | 32 In this table you can see that the rule is Input x2 = Output Hope This helped!
What output is 4 more than twice the input 3?
10
How do you do in and out in math?
Assuming by in you mean input and out you mean output. Input is the value that goes in while the output is the value you receive. Between these terms is a rule, called the nth term that will always work to help you find the input/output. For example. Our input is 2, and our output is 10 the rule here could be the input multiplied by 5 equals the output, or it can be something extremely difficult and unfathomable even to a banker...
What sentence expresses the pattern of Fn if a table shows some input and output values for a function named Fn?
The output is 1 more than 10 times the input.
What will be input and output current of 10 KVA 3 phase UPS?
A 10 KVA 3-phase UPS will have an input and output current that depends on the specific voltage of the system. You can calculate the current by dividing the apparent power (in this case 10 KVA) by the square root of 3 multiplied by the voltage. For example, for a 208V system, the input and output current would be approximately 28.8 amps.
