you need to move the graph to the left, which means that the thing goes inside the bracket ( f(x+something) ). You want to move it to the left by the middle value, which can be determined by the average of both p and q, so you do (0.5x(p+q)). Therefore the answer is f(x+0.5(p+q)).
you need to move the graph to the left, which means that the thing goes inside the bracket ( f(x+something) ). You want to move it to the left by the middle value, which can be determined by the average of both p and q, so you do (0.5x(p+q)). Therefore the answer is f(x+0.5(p+q)).
This is the answer to ii.
Thanks but the answer to ii) is y=f(x - ((a+b)/2)). This is shown in the mark scheme which I have attached in my first post.
(ii) I certainly had to take a look at the answers to figure out what they even mean by part (ii) but once you're happy that the middle of the 'dip' occurs half-way between t=a and t=b, then the translation should be obvious as you want this bit to occur at t=0. As this is a periodic function, it doesn't matter whether you do f(t+2a+b) or f(t−2a+b)
(iv) The amplitude is y, so increasing it by k means moving it up by k hence y↦(y−k)
(v) Yes it would be what you say
(b) Just sketch 4 different graphs, applying transformation after transformation, and you'll get to the last one.
EDIT: I just realised their graph axes don't make sense either. It should be f(t) on the vertical, and y=f(t) in the context.
(ii) I certainly had to take a look at the answers to figure out what they even mean by part (ii) but once you're happy that the middle of the 'dip' occurs half-way between t=a and t=b, then the translation should be obvious as you want this bit to occur at t=0. As this is a periodic function, it doesn't matter whether you do f(t+2a+b) or f(t−2a+b)
(iv) The amplitude is y, so increasing it by k means moving it up by k hence y↦(y−k)
(v) Yes it would be what you say
(b) Just sketch 4 different graphs, applying transformation after transformation, and you'll get to the last one.
EDIT: I just realised their graph axes don't make sense either. It should be f(t) on the vertical, and y=f(t) in the context.
Thank you so much -this is starting to make much more sense.
iv) The amplitude is y, so increasing it by k means moving it up by k hence y↦(y−k)
I am a bit confused here with the answer you have given because the mark scheme answer is y=f(x) +k.
I know the amplitude of the signal is the height of the signal because it says so in (i)
The mark scheme answer y=f(x) +k doesn’t make much sense to me either because doesn’t this just shift the signal graph up instead of increasing the height/peak of the signal by k?
(ii) I certainly had to take a look at the answers to figure out what they even mean by part (ii) but once you're happy that the middle of the 'dip' occurs half-way between t=a and t=b, then the translation should be obvious as you want this bit to occur at t=0. As this is a periodic function, it doesn't matter whether you do f(t+2a+b) or f(t−2a+b)
(iv) The amplitude is y, so increasing it by k means moving it up by k hence y↦(y−k)
(v) Yes it would be what you say
(b) Just sketch 4 different graphs, applying transformation after transformation, and you'll get to the last one.
EDIT: I just realised their graph axes don't make sense either. It should be f(t) on the vertical, and y=f(t) in the context.
Thank you so much -making much more sense.
iv) The amplitude is , so increasing it by means moving it up by hence
I am a bit confused here with the answer you have given because the mark scheme answer is y=f(x) +k.
I know the amplitude of the signal is the height of the signal because it says so in (i)
The mark scheme answer y=f(x) +k doesn’t make much sense to me either because doesn’t this just shift the signal graph up instead of increasing the height/peak of the signal by k?
iv) The amplitude is , so increasing it by means moving it up by hence
I am a bit confused here with the answer you have given because the mark scheme answer is y=f(x) +k.
I know the amplitude of the signal is the height of the signal because it says so in (i)
The mark scheme answer y=f(x) +k doesn’t make much sense to me either because doesn’t this just shift the signal graph up instead of increasing the height/peak of the signal by k?
But isn't that what shifting up the graph by k does? Increase the peak by k?
So then as I said, you replace y by y−k hence y−k=f(t)⇒y=f(t)+k
So is the peak/amplitude measured from the x axis and not from the lowest point of the graph?
If you at the definitions: https://en.wikipedia.org/wiki/Amplitude I believe in this question they are using the Peak Amplitude rather than Peak-to-Peak, so they are measuring from the t-axis to the highest point.
TBH I'm not comfortable with this question and the way its worded/worked out.
If you at the definitions: https://en.wikipedia.org/wiki/Amplitude I believe in this question they are using the Peak Amplitude rather than Peak-to-Peak, so they are measuring from the t-axis to the highest point.
TBH I'm not comfortable with this question and the way its worded/worked out.