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Sample mathematical exercises for engineering
Professor John H. Davies
2012 July 5
Here are a few exercises to help you to revise your mathematics before you come to univer- sity. All of the techniques should have been covered in Higher Mathematics but the questions are dressed up in the language of engineering, which may make them a lot more challenging! However, they will provide a good introduction to studying at university. The examples are from electronics because you may have encountered some of the material in Higher Physics at school. You will see plenty of applications to your discipline when you arrive here. Please don’t get the idea that the curriculum is dominated by mathematics: it is definitely engineering. However, professional engineers use mathematics as a tool to help them solve problems, which means that you must be able to do basic calculations quickly and reliably – almost automatically. You won’t be able to concentrate on the engineering if it takes you half an hour to solve a quadratic equation, for instance. We will help you to develop this skill during your university course. It will be a challenge but very profitable for your future career. I have included numerical answers to some of the questions. Full solutions are available on the School of Engineering’s web site at www.gla.ac.uk/engineering/infopacks. Please don’t look at these until you have tried the exercises yourself.
- Figure 1(a) shows a widely used circuit called a potential divider formed by two resistors. The input and output voltages are given in terms of the resistances by
Vout =
R 2
R 1 + R 2
Vin.
Use this to find the unknown quantities in figures 1(b)–(e). [0.5 V, 500 Ω, 12 V, 16 kΩ.]
V in
V out
R 1
R 2
(a)
5 V
V out
9 kΩ
1 kΩ
(b)
10 V
0.4 V
12 kΩ
(c)
R (^2) 2 V
5 kΩ
(d)
1 kΩ
V in
1 V
(e)
2kΩ
R 1
9 V
Figure 1: A selection of potential dividers.
(a) inverting amplifier
−
R 1 R 2 V out
V in
(b) non-inverting amplifier
−
R 1
R 2 V out V in
Figure 2: The classic inverting and non-inverting amplifier circuits.
- A remote control draws 10 mA while it is being used and 10 μA when it is idle. (Make sure that you know the powers of 10 for the prefixes in mA and μA. How about kA and nA?) What is the average current drawn, assuming that it is used for 5 minutes per day? Which is more significant, the current drawn when it is operating or idle? [45 μA] The control’s batteries are rated at 100 mAh. This means that the product of the current in mA and lifetime in hours is 100. For example, they will provide 100 mA for 1 hour or 0.1 mA for 1000 hours. How long will they last in the remote control? [About 3 months]
- Figure 2 shows two classic circuits based on an operational amplifier – the component shown by the triangular symbol. They act as (a) inverting and (b) non-inverting ampli- fiers. You may have come across these circuits before but don’t worry if you haven’t because you only have to do the mathematics here, not the electronics! Standard circuit analysis leads to these equations for each circuit.
(a)
Vin R 1
Vout R 2
= 0 , (b)
−Vin R 1
Vout −Vin R 2
Rearrange each of these equations to find expressions for Vout/Vin, called the voltage gain of the circuit. Watch the signs! What would the gain be in each case if R 1 = 1 kΩ and R 2 = 10 kΩ? [−10, 11]
- When designing a circuit, a resistance R is required to obey the equation
R^2 − 2 R 1 R + R^21 − R^22 = 0
where the values of R 1 and R 2 are known. What sort of equation for R is this? Find an algebraic expression for the possible values of R. If R 1 = 10 Ω and R 2 = 30 Ω, find the value of R. Why is there only one? [40 Ω.]
- A signal with voltage V (t) is processed through two circuits. The first adds a constant 5 V and the second squares its input. Find an expression for the final output Vout(t). What is the output when the input is V = 1 V? [36 V]
Solution. We just have to rearrange the given equation in the four possible ways to get the different unknown quantities. This isn’t necessary for (b) because the equation gives Vout directly. To get the others it is a good idea to multiply throughout by (R 1 + R 2 ), which gets rid of the fraction: R 1 Vout + R 2 Vout = R 2 Vin.
I’ll start from here to solve (c)–(e). For (c) we need to find R 2. Take the term R 2 Vout from the left to the right-hand side, changing the sign. R 1 Vout = R 2 Vin − R 2 Vout.
Now take out the common factor of R 2.
R 1 Vout = R 2 (Vin −Vout).
Finally, divide both sides by (Vin −Vout).
R 1 Vout Vin −Vout
= R 2.
For (d) we need to find Vin. Take out the common factor of Vout on the left.
Vout(R 1 + R 2 ) = R 2 Vin.
Divide by R 2 and we are there. Vout(R 1 + R 2 ) R 2
= Vin.
Finally, (e) needs R 1. This is almost the same as finding R 2 down to the line
R 1 Vout = R 2 (Vin −Vout).
This time, divide both sides by Vout.
R 1 =
R 2 (Vin −Vout) Vout
(a) inverting amplifier
−
R 1 R 2 V out
V in
(b) non-inverting amplifier
−
R 1
R 2 V out V in
Figure 2: The classic inverting and non-inverting amplifier circuits.
- A remote control draws 10 mA while it is being used and 10 μA when it is idle. (Make sure that you know the powers of 10 for the prefixes in mA and μA. How about kA and nA?) What is the average current drawn, assuming that it is used for 5 minutes per day? Which is more significant, the current drawn when it is operating or idle? [45 μA] The control’s batteries are rated at 100 mAh. This means that the product of the current in mA and lifetime in hours is 100. For example, they will provide 100 mA for 1 hour or 0.1 mA for 1000 hours. How long will they last in the remote control? [About 3 months]
Solution. Prefixes: k = kilo = 10^3 , m = milli = 10−^3 , μ = micro = 10−^6 , n = nano = 10−^9 , p = pico = 10−^12. The average current is given by
Iaverage =
Ioperating × Toperating + Iidle × Tidle Toperating + Tidle
This is much the same as finding the average velocity for a journey. Let’s use μA for all currents and minutes for all times. There are 60 × 24 = 1440 minutes in a day and the remote control is used for 5 minutes so it is idle for 1435 minutes.
Iaverage =
10 000 × 5 + 10 × 1435
= 35 + 10 = 45 μA.
This shows that the average current is made up of 35 μA from the time when the remote control is operating and 10 μA when idle. Thus the operating current is more significant although the system spends most of its time idle.
- Figure 2 shows two classic circuits based on an operational amplifier – the component shown by the triangular symbol. They act as (a) inverting and (b) non-inverting ampli- fiers. You may have come across these circuits before but don’t worry if you haven’t because you only have to do the mathematics here, not the electronics! Standard circuit analysis leads to these equations for each circuit.
(a)
Vin R 1
Vout R 2
= 0 , (b)
−Vin R 1
Vout −Vin R 2
Solution. There are several ways of solving the quadratic equation for R. I think that the best way is to spot that R^2 − 2 R 1 R + R^21 = (R − R 1 )^2. This is the standard result for the expansion of (A − B)^2. Then (R − R 1 )^2 − R^22 = 0.
Take R^22 over to the right-hand side.
(R − R 1 )^2 = R^22.
Now take the square root of both sides. The only tricky point is the sign. Remember that the square roots of A^2 are ±A. Here we should put a ± sign on both side but there are really only two possibilities: either the two sides have the same sign or the opposite signs.
R − R 1 = ±R 2.
Finally, take R 1 across to get the solutions.
R = R 1 ± R 2. There are plenty of other ways of solving this. Completing the square gives exactly the same working. You can also use the standard formula for the roots of ax^2 + bx + c = 0. Here x = R, a = 1, b = − 2 R 1 and c = R^21 − R^22. The standard formula is
x =
−b ±
b^2 − 4 ac 2 a
Substitute our expressions into this.
R =
2 R 1 ±
(− 2 R 1 )^2 − 4 × 1 × (R^21 − R^22 )
2 × 1
Simplify this.
R =
2 R 1 ±
4 R^21 − 4 (R^21 − R^22 )
2 R 1 ±
4 R^21 − 4 R^21 + 4 R^22
2 R 1 ±
4 R^22
2 R 1 ± 2 R 2
= R 1 ± R 2.
It is a lot less effort to spot the pattern (A − B)^2! The numerical values give 10 ± 30 Ω = 40 Ω or − 20 Ω. The second result is negative and you can’t buy negative resistors!
- A signal with voltage V (t) is processed through two circuits. The first adds a constant 5 V and the second squares its input. Find an expression for the final output Vout(t). What is the output when the input is V = 1 V? [36 V]
Solution. The output is Vout = (V + 5 )^2 because the addition is done first, then the squaring. (It would be Vout = V 2 + 5 if the operations were done in the reverse order.)
